Binomial Theorem Class 11
Master Binomial Theorem Class 11 with comprehensive NCERT Solutions, Practice Questions, MCQs, Sample Papers, Case Based Questions, and Video lessons.
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Binomial Theorem Class 11 – NCERT Solutions
Each question below opens its complete step-by-step Teachoo solution.
Ex 7.1
14 questionsEx 7.1,1
Ex 7.1,1 teachoo.com
Expand the expression (1 — 2x)>
(1- 2x)
We know that
(a+b)? ="C, a 4+"C, ah tbt4"C, a2 bet tC, at be 44°C, be
Hence
(a+b) = 5C, a® +°C,atb? + 5C, a? b? + 5C, a2b? + °C, ab* + °C, b®
_ sl 5 5! ant 5! 31h?
“oso? then? b hep b
5! 213 5! a 5] 5
*acs-a? b + Gia ai * 5s 3)! b
Ex 7.1,2
Ex 7.1, 2 teachoo.com
5
; 2 x
Expand the expression G _ *)
We know that
(a+ b)P="Cya"+"C, a"~1bt+"C, a0? b? +0 +t Cat bh t+"C, b"
Hence
(a+b)? = 5c, a5 +5C, ab! +5C, ab? +5, a2b? +5C,ab4 +5C, b>
__ 5! 5 5! ant 5! 312
“ons—o ? * TG b * os pi? b
5! 213 5! a 5! 5
*3as-a 2 b + as miab * Sus-5)! b
-_5!_ os Sd yt a8 2g 2h gp a ht gg DS
- oxi? *txa? b+ 34 b +32 4 b +a ab + FoiP
Ex 7.1,3
Ex 7.1, 3 teachoo.com
Expand the expression (2x — 3)®
(2x - 3)°
We know that
(a+b)? ="Cya"4+C, a" 1bt+"C, al? b? +00 FC, ath 1 4+-"C, b"
+°C,a°b”
Hence
(a+b)®= 6¢, a +6C,a5 b! + 5C, atb2 + °C, a3 b? + 8C, a2 b4
+°Coab> +°C, b®
_ 6! 6 6! 5 6! aw 6! 343
“oreo? * Te-D!? b+T6-mi@ b *ye-3!° b
6! aha 6! 5 6! 6
*ie-4)? b + reesei + e1e-Hie
Ex 7.1,4
Ex7.1,4 teachoo.com
ony
Expand th (+2)
xpand the expression (2 + —
We know that
(a+b)? ="Cy a" +C, a" tbt4+C, al? b? 4 Cp ate 14°C, bo
Hence
(a +b) = 5C, a° +°C, ath? +5C,a%b? +°C, a2b? +°C,ab* +5C,b>
_ sl 5 5! ant 5! 312
ONS —o 4 *hG-p? b *aG-m4 b
5! 13 5! a 5! 5
*3G-D! a’b +a mi2b + 51(5 —5)! b
Ex 7.1, 5
Ex 7.1,5 teachoo.com
6
1
Expand (x +=)
We know that
(a+b)? ="C,a"4+"C, a" 1bt+"C, al? b? +0 Cath 1 4+"C, b"
Hence
(a+b)&= °C, a& +°C,a> bt +°C,atb? +°C,a2b? +°C,a? b*
+°C, a b> + °C, b®
= © 46 8 as py yon pr © on 53
oreo? x1 né-p!4 b+Te-mi4 b *3ue-3)° b
6! aha 6! 5 6! 6
+ne-a b + Fe- 5 ae + sie-Hie
Ex 7.1, 6
Ex 7.1, 6 teachoo.com
Using Binomial Theorem, evaluate (96)?
(96)? = (100 - 4}?
We know that
(a+b)? ="Cya"+"C, a" 1bt+"C,a"-72 b? +t Cp at bh 14°C, b"
Hence
(a+b)? = °C, a? b°+ °C, a*b* + °C, a" b? + °C a° b*
3! 3 3! > 3! > 3! 3
-aG-o xI+7G-p4 b+7G@opi ab *zGopit*b
3! 3x2! 3x2! 3!
sy at a b+ ab?+ Fa?
=a? + 3a’b + 3b2a + b?
Ex 7.1,7
Ex 7.1, 7 teachoo.com
Using Binomial Theorem, evaluate (102)°
(102)° = (100 + 2)°
We know that
(a+ b)P="Cya"+"C,a"-tbt+C, ah? b? + tC, athe t+ ec, be
Hence
(a+ b)S= °C, a + °C, a*b? +5C, a? b* +5C, a2b? +5C,ab* +°C;b°
-_ >! 5 Ss! 4b1 5! 3 2
ogo" * 1 *he-p? b tae b
5! 213 5! a 5! 5
*3G-a)! a’b +aG@—mia> * 55 — 5)! b*x4
Ex 7.1,8
Ex 7.1, 8 teachoo.com
Using Binomial Theorem, evaluate (101)*
(101)* = (100 + 1)*
We know that
(a+ b)D="Cyah+PC a" tht +"C, a2 bt wt MC) att 40°C, be
Hence,
(a+b)* =4C, a* + 4C, a? bt + 4C, a b* + 4C, atb? + 4C, bt
al a 4 3h 41 ere 4 4
~ OKa— oy? *Txa-p? b *pa-pi" b *ha-a1" b + haa?
= ats ab + a? bp? + ab? + bt
1x4! 1x 3! 21x 2! 31x 1! 4! 0!
Ex 7.1,9
teachoo.com
Ex 7.1,9
Using Binomial Theorem, evaluate (99)>
(99)> = (100 - 1)°
We know that
(a+b)? ="Cy a" +C, a" tbt4+C, al? b? 4 Cp ate 14°C, bo
Hence
(a+b) = 5, a° +°C, a*bt +5C, a? b* +5C, a2b? +°C, abt + °C, b®
_ si 5 5! at 5! 312
~oONs—o 4 *hG-pi4 b *aG-m4 b
5! 13 5! a 5! 5
+*3G-D! a’b +a mi2b + 51(5 —5)! b
Ex 7.1,10
Ex 7.1, 10 teachoo.com
Using Binomial Theorem, indicate which number is larger (1.1)19°° or
1000.
(1.1)10000 = (1 + 0.1)10000
We know that
(a+ by"="Cya"+"C,a"-tbht+ "Ca" bt wt MC) pat br-24"C, be
Hence
(a + b)10000 = 10000¢ qi00004 10000¢ 9999 bt HF iscccusseosace
= 10000 10000 4 4 10000! a s0098 4
01(10000 — 0)! 11(10000 — 1)! vere
= 100001 10000 x 1 4 10000 x 9999! 5009 b
1 x10000! 11x99991 sesenenesnenee
= a000 + 10000a9999 b t.. ec eeseee
Ex 7.1, 11
Ex 7.1, 11 teachoo.com
4 4
Find (a + b)*-(a—b)*. Hence, evaluate (V3 + V2) -(V3-—V2) .
We know that
(a+ b)P="Cya"+"C, a" 1bt+"C, al? b? +0 tC, ath t+"C, b"
Hence,
(a+b)*=4C, a*+4C, a2 b+ 4C, a2b? + 4C, atb?+ 4C,b*
_ 4, 4! 3h4 4) on 4! ogi,, 4h og
“ao? tixa-pi? b *7a-pi" b *ha-ai" b * haa) b
= 44,4! 23 A ope, 34" pa
= eat gi bt gab? + TG abt ab
=a‘ +4a*b + 6a?b* +4 ab?+b*
Ex 7.1,12
Ex 7.1, 12 teachoo.com
6
Find (x + 1)°+ (x-—1)®. Hence or otherwise evaluate (v2 + 1) +
6
(v2-1).
We know that
(a+b)? ="Cya"4+"C, a" 1bt+"C, al? b? +00 Cg ath 1 4+"C, b"
Hence
(a+ b)&= °Cy a® +°C,a° bt +°C, ath? +°C,a2b? + °C, a? b*
+°C, ab + °C, b®
= © a6 8 as py ont pr __ © on 3
~oe-o* * 1+ né-p!4 b+Te-mi4 b *3ue-31° b
6! aha 6! 5 6! 6
+ne-a b* + rae Than * o6-0! b
Ex 7.1,13
Ex 7.1, 13 - Introduction teachoo.com
Show that 9"! — 8n —9 is divisible by 64, whenever n is a
positive integer.
Numbers divisible by 64 are
64=64x1
128 =64x2
640 = 64x 10
Any number divisible by 64 = 64 x Natural number
Hence, In order to show that 9°"! - 8n - 9 is divisible by 64,
We have to prove that
9"™1 _ 8n -9 = 64k, where k is some natural number
Ex 7.1,14
Ex 7.1, 14 (Method 1) teachoo.com
n
Prove that » 3rnc, = 4"
r=0
By Binomial Theorem,
nm
yc geo rpr= (at+b)"
Tr
r=0
Putting b = 3 and a = 1 in the above equation
n
» nC, ye73r = 6 + 3)n
r=0
nr
Y 3rmc, = (4)
r=0
Hence proved
Examples
17 questionsExample 1
Example 1 teachoo.com
Expand (x? + 2" #0
xpand (x? + =} , x
We know that
(a+ b)P="Cya"+C, a" 1bt+"C, a"? b? + tC), atb-1+C, be
Hence,
(a+b)*=%C, at +4C, a? b+ 4C, a2 b? + 4C, ab? + 4C, b*
4 a Al 712 Al 3 4l a
= o(4-0!¢ *Ta-p? b + han *fa-4) b
_ 44, 4g 4! ooo, 4h, 4h og
"Txa* tix? b+ ona b aren ab +x Tika
=a‘ + 4a°b + 6a? b? + 4 ab? + b*
Example 2
Example 2 teachoo.com
Compute (98)°.
(98)5 = (100 — 2)5
We know that
(a+ by? ="Cya"+"C, ah tbht+"C, a2 bt + Cathe 14+", be
Hence
(a+ b)>=
5Cy a> +9C, a* bb? + 5C, a? bb? + 5C, a2b? + 5C, a b*+ °C, b®
_ sl 5 5! ant 5! 312
~ Oso! 4 *tG-p4 b *nG-pi4 b
5! 13 5! a 5! 5
*3G-D! a’b +a i2b * 51( 5-5)! b
Example 3
Example 3 teachoo.com
Which is larger (1.01)*909° or 10,000?
(1.01)1000000 = qa + 0.01)1000000
We know that
(a+b)? ="Cya"4+C, a" tbt4"C, a"? b2 4G, atb" 44 "C, be
Hence
(a + fy) 1000000 = 1000000 qio00000 b? + 1000000¢ 799999 bt Ficccccessesee
= 1000000 1000000 4 4 1000000! oon000p 4
01(1000000 — 0)! 11(1000000 — 1)!
= 1000000! 1000000 4, 1900000 x 999999! socsa0 py
1 x 1000000! 1x 999999!
= 41900000 4 1000000a2999% b +
Example 4
Example 4 (Introduction) teachoo.com
Using binomial theorem, prove that 6"—- 5n always leaves remainder
1 when divided by 25.
Taking examples
264
25 25
we can write 26 as 25x1+1
Stig td
25 25
we can write 51 as 25x2+1
251 1
25 7 1055
we can write 251 as 25x 10+1
Question 1
Example 5 teackoo.com
Find a if the 17th and 18th terms of the expansion (2 + a)°° are equal.
We know that
General term of expansion (a + b)"is
Tha = "C, a" bt (1)
Finding 17 term Finding 18" term
Taz = Tiga Of (2 + a)? Tag = Taya1 of (2 + a)°°
Putting r=16,n=50, a=2 Putting r=17,n=50, a=2
and b =a in (1) and b =a in (1)
Taos 1 = °C (2)5-19.. (a)#6 Taper = 5C yy (2}9-17, (alt?
Tyz = Cag - (2)** . at? Tag = Cy, (2)79. (a)?
Question 2
Example 6 teackoo.com
Show that the middle term in the expansion of {1 + x)?" is
aur) 2" x", where n is a positive integer.
Given
Number of terms = 2n which is even
So,
. 2n th
Middle term = (+ 1)" term
=(n+1)"term
Hence, we need to find T, , ,
Question 3
teachoo.com
Example 7
Find the coefficient of x°y? in the expansion of (x + 2y)?.
We know that
General term of expansion (a + b)"is
Th = °C, ant br
For {x + 2y)°,
Puttingn=9,a=x,b=2y
Trea = 7C, (x7! (2y)'
=9C, (JF (yy (2) (1)
We need to find coefficient of x® y?
Question 4
Example 8 teachoo.com
The second, third and fourth terms in the binomial expansion (x + a}"
are 240, 720 and 1080, respectively. Find x, a and n.
We know that general term of (a + b)" is
Thar = "C, (ayer. (by (a)
Given that second term of {x + a)" is 240
i.e. T, = 240
Tai = 240
Putting r=1,a=x&b=a
Tyia = 7, (x) (a)*
T,="C, xa
240="C.x™la (2)
Question 5
Example 9 teachoo.com
The coefficients of three consecutive terms in the expansion of {1 + a}"
are in the ratio 1: 7 : 42. Find n.
Let the three consecutive terms be (r— 1)", rtY and (r + 1)" terms.
Le. Teo T & Tea
We know that general term of expansion (a + b)" is
Traie "C, a -'bhr
For (1+a)",
Puttinga=1,b=a
That = "C, yn-e a’
Tra = "C, at (1)
+. Coefficient of (r + 1)" term = °C,
Question 6
Example 10 teachoo.com
6
Find the term independent of x in the expansion of Gx? - =) »x> 0]
Calculating general term
We know that general term of expansion (a + b)" is
Tea = *C, (a)"™(0)"
3 9 146
For general term of expansion Gx - =)
Putting n=6,a=2x?,b=-—
2 3x
3 6-97 7_4\r
Tar, (Gx) (Z)
Question 7
Example 11 teachoo.com
If the coefficients of a'~ 1, a’ and a’*1 in the expansion of (1 + a)"
are in arithmetic progression, prove that n? — n(4r + 1) + 4r? - 2 =0.
We know that
General term of (a + b)"
Test = "Cart br
For (1+ a)"
Puttinga=1&b=a
Tear = C-(1)9"". at
Trea = "Cat (1)
Hence, Coefficient of a‘ = "C,
Question 8
Example 12 teackoo.com
Show that the coefficient of the middle term in the expansion of
(1 +x)?" is equal to the sum of the coefficients of two middle
terms in the expansion of (1 + x}?"~1.
Middle term of (1 + x)?"
Since 2n is even,
th
Middle term = e +1)
=(n+1)" term
= Tht
We know that general term of expansion (a+ b)" is
Tra1= nC, a"-F bt
Question 9
Example 13 teackoo.com
Find the coefficient of a* in the product (1 + 2a)* (2 — a)? using binomial
theorem.
We know that
(a+ b)"="Cya"4"C, a" tht 4. + 9C,_, at b "4 °C, b"
Hence
(a+b)* =4C, a4 + 4C, a2 b? + 4C, a? b? + 4C, atb? + 4C, a°b*
_ 4, 43 4 45, 4 3 4 oy
“aa-o? “ne-»? b yal" b * haa ab tFa@-a
= ate Hath + a? bp? + ab} + pt
1x4! 3! 21x 2! 31x14 1! 0!
= at ¢ XH aap g 3X4 oo pr 4g 233 yt eps
3! 2x2! 3! 4aPx1
=a‘ + 4a*b + 6a*b* + 4 ab? + b*
Question 10
Example 14 teachoo.com
Find the r‘* term from the end in the expansion of (x + a)".
We know that
(at+b)"= "C, atb°+ °C, abt 4. OC, (aj be t+ "Cc, ao be
= al $C, abt t eee t "Cy atb™? +b?
From a uN u— wv
__ First Second nth (n+ yin
starting
Term Term Term Term
=b™ tC, at bt tet MC, att bt + at
Ww a | uss u
A
Fromend fi second nv (n+ 1}
Term Term Term Term
Question 11
Example 15 teackoo.com
. . . . 3 1 \18
Find the term independent of x in the expansion of (Vx + =) ,
x>0.
Calculating general term of expansion
We know that general term of (a + b)" is
Tra = "CG, (a). (a)?
; 3 4 \18
For general term of expansion (Vx + =m)
. _ 3 1
Putting n=18, a= Vx, b=s95
; 18¢ (2/yy18-r (_1_\"
+ Thar BC, (Vx) (=)
Question 12
Example 16 teachoo.com
The sum of the coefficients of the first three terms in the expansion
of (x-5)" ,x #0, m being a natural number is 559. Find the term
of the expansion containing x°.
We know that
General term of expansion (a + b)"is
That = nC, a"-' ht
3
For (x-=)"
. =3
Puttingn=m,a=x,b=—>
= (23V"
Question 13
Example 17 teachoo.com
If the coefficients of (r —5)"" and (2r— 1)" terms in the expansion of
(1 + x)*4 are equal, find r.
We Know that
General term of expansion (a + b)" is
T.41=%C,a™ bi!
General term for (1 + x)**
Putting a=1,b=x,n=34
Teg = tC, 10 x!
Thai = MC, x! (1)
Miscellaneous
10 questionsMisc 1
Mise 1 (Introduction) teachoo.com
lf a and b are distinct integers, prove that a — b is a factor of a" —b”,
whenever n is a positive integer. [Hint: write a" = (a—b +b)" and
expand]
As 4 divides 24,
4is a factor of 24
We can write
24=4x6
Similarly,
If (a—b) is a factor of a® — b"
then we can write
a"- b"=(a-b)k
where k is a natural number
Misc 2
Misc 2 teachoo.com
Evaluate (v3 + V2 )°-(v3-v2)°.
Finding (a + b)°— (a — b)®
We know that
(a+ b)P="Cya"+"C, a"-1bt+"C, al? b? +0 tC, at bh t+"C, b"
Hence
(at+b)®= §C, a® +°C,a°b! +°C,atb? +°C,a2b? + °C,a2b*
+ °C, ab + °C, b®
= © _ 46 Fas py 8 nt pr ns 3
oreo! 4 x1+ né-pi4 b+Te-mi4 b *ze-a° b
6 naps pS gs 5 9! eps
*ne-a b + 5ge— =a * eie-oie
Misc 3
Misc 3 teachoo.com
Find the value of (a2 + Va? — 1)4+(a2-vVa? —1)*.
We know that
(a+ b)P="Cya"+"C, a" tbt+"C, ah? b? + tC, at bet +c, bo
Hence,
(a+b)*=4C, a4 +4C, a3 b1 + 4C, a2 b? + 4C, atb?+4C,b*
_ 4g 4 3h 4) ere 4 4
tao! *ixa-p? b +7a-pi b +a ai b + aa—pi
=o at4 3p + ab? + ab? + bt
1x4! 1x 3! 21x 2! 31x41! 4! 0!
=a‘ + 4a? b + 6a? b? + 4 ab? + b*
Misc 4
Misc 4 teachoo.com
Find an approximation of (0.99) using the first three terms of its
expansion.
(0.99)5 = (1-0.01)5
We know that
(a+b)? ="CyaP+"C, at bt + "C, ah 2b? to. +9C,_, at" 1+°C, be
Hence
(a+b)= 5¢, 95 49C, atbt + 5C, a2b? +5C, ab? +°C,ab* +5C, b®
=95 5! 4,1 5! 3 2 5! 23
Fat to-pi4 b *ne-m b *he-a? b
5! ans
+a - aah +b
Misc 5
Misc 5 teachoo.com
x 2 4
Expand using Binomial Theorem (1 + 77 =) X#0.
We know that
(a+ b)P="Cya"+"C, a"~1bt+"C, a"? b? +0 tC, ath t+"C, b"
Hence,
(a+b)*=4C, a*+4C, a2 b+ 4C, a2b? + 4C, atb?+ 4C,b*
_ 4, 4l 3h4 4 ere 4l a
-oa-01° Tixa-pi? b *7a-p!" b *ha-a4 b cers b
- 1 34,_4' 23 Leer) 4 34" pa
= Txa* tia b+ a4 b + 3x ab +a oi P
=a‘* + 4a?b+6a2b?+4ab?+b* ...(1)
Misc 6
Misc 6 teachoo.com
Find the expansion of (3x? - 2ax + 3a’)? using binomial theorem.
We know that
(a+b)? ="Cya"4+"C, a"-1bt4+"C, al? b? +0 Cath 14+"C, b"
Hence
(a+b)? = °C, a? +3C, a2b* + °C, a b* + 3C; b?
=42 3 2 3} 2 3
=a *Te-p:? b+ Gp ab* +b
=a? + 3a*b + 3b*a+b? ...(1)
Question 1
Misc 1 teachoo.com
Find a, b and nin the expansion of (a + b)" if the first three terms of th
expansion are 729, 7290 and 30375, respectively.
We know that
(at b)"="C,a"+"C, a" 1 bt +"C,a"~? b? +. FC, at b "1+ °C, bY
= a4", a" bts "Ci ah 2b tu eMC) at bt t+ bY
So first 3 terms are a", "C, a"~*b and "C, a"~? b?
Also, it is given that their value are 729,7290 and 30375
o am=729 (1)
"C,a"-*b = 7290 (2)
"C, a"-2b2 = 30375 (3)
Question 2
Misc 2 teachoo.com
Find a if the coefficients of x? and x? in the expansion of (3 + ax)?
are equal.
We know that
General term of expansion (a + b)" is
Tha =", (a)?-". bt
For (3 + ax)?,
Puttinga=3,b=9x &n=9
General term of (3 + ax)? is
Thea = °C, (3)9-". (ax)
=9C, 39>" ale xt (1)
Question 3
Misc 3 teachoo.com
Find the coefficient of x° in the product (1 + 2x)® (1 — x)’ using binomial
theorem.
We know that
(a+b)? ="Cyat+C,a"-tbt + wi. + C, at bet 4"C, be
Hence
(a+ b)®= ©C,a®+®C, a> bt + °C, at b? + °C, a3 b? + °C, a? b*+ °C, ab° + °C, b®
= 36 6 5 6 4 2 6! 3 3
=a + Te-p!? b+a@—mi? b *Se-ai? b
6! aha 6! 5. he
*ne-ae b + Ties a? +b
= abe a p+ — ath? + a3 b? + a? bt
1x5! 2!x 4! 313! 4!2!
+—_ab’+ bo
5Ix1
Question 4
Misc 8 teachoo.com
Find n, if the ratio of the fifth term from the beginning to the fifth term
+ . 4 1\",
from the end in the expansion of (V2 + a) isV6:1
V3.
We know that
General term of expansion (a + b)"
That = "C, abr
Fifth term from beginning
We need to calculate T, =T,,,
+ _ _4 _ ==
Puttingr=4, a= V2, b=(z-)
4 _ 1
Tq. "C,(¥2)" “(z)
General Term of Binomial Theorem
12 questionsQuestion 1
teackoo.com
Ex 8.2, 1 - Introduction
Find the coefficient of x° in {x + 3)®
In 18x
|
Coefficient
Coefficient of x° = 18
Question 2
Ex 8.2, 2 teachoo.com
Find the coefficient of a>b7 in (a — 2b)?
We know that general term of expansion (a+ b)"
Tras = °C, (a). bt
For general term of expansion (a — 2b }'2
Putting n=12,a=a,b=—2b
Trex = 2C, (a}?-#. (2b)!
=%C, (a)2—F. 2)". (by
= 6, (-2) (ay?-"(b)" (2)
We need to find coefficient of a°b”
Question 3
Ex 8.2, 3 teachoo.com
Write the general term in the expansion of (x? — y)®
We know that
General term of expansion (a + b)"is
For (x? —y)®
Puttingn=6,a=x?,b=-y
Trea = °C, 0°)°- (-y)'
= 6! 26-r) f_4\r r
a OO AY)
-i-1\ 6! 12 -2r r
=Ch rien ™ Y
Question 4
Ex 8.2, 4 teachoo.com
Write the general term in the expansion of (x? — yx)!?, x #0
We know that
General term of expansion (a + b)"is
Tha = "C, a" br
For (x? — yx),
Putting n=12,a= x?, b=—yx
Thea = 7C,07}7-". (— yx)
= RC, (x)242-9) . (-1)'- y ox
= nO x24-2r. (-1)": y Je
Question 5
Ex 8.2, 5 teachoo.com
Find the 4 term in the expansion of (x — 2y)!?.
We know that
General term of expansion (a + b)"is
Tut = "C, a br
We need to find fourth term
i.e. T, = T;,,0f expansion (x — 12y)12
Puttingr=3,a=x,b=-2y,n=12
T3441 7C, (x)? 73 (-2yp
Ex 8.2 6
Ex 8.2, 6 teachoo.com
. th . . 1 \18
Find the 13° term in the expansion of (ox - =R) X#0.
We know that
General term of expansion (a + b)" is
We need to calculate 13" term
. . 1 \'8
(ie. T,3= Ty, ) of expansion (ox - =a)
. 1
Putting r=12,n=18,a=9x&b =a
_4\12
Tyan = 8Cq2 (9x8? (=)
Question 7
Ex 8.2, 7 teachoo.com
. . . . \7
Find the middle terms in the expansions of (3 - =)
Number of terms = n = 7
Since n is odd there will be two middle termx
th th
C=) term = Cj) term = 4" term
2
th th
& Ce + 1) term = ‘ee + 1) term = {4+ 1)"=5" term
Hence we need to find 4" and 5" term
i.e. T, and T,,
We know that general term of (a+ b)" is
Question 8
Ex 8.2, 8 teackoo.com
x 10
Find the middle terms in the expansions of G + oy)
Number of terms n = 10
Since n is even.
There will be one middle term
Middle term = > +1
= 44
2
=5+1
= 6" term.
Hence we need to find 6“ term. i.e. T-
Question 9
teachoo.com
Ex 8.2, 9
In the expansion of (1 + a)™*", prove that coefficients of a™ and a”
are equal.
We know that
General term of expansion (a + b)" is
For (1+a)™*",
Puttingn=m+n,a=1,b=a
Trt = n+mc, (1jyotmr (a)'
=nrmc, (a) (1)
Question 10
Ex 8.2, 10 teachoo.com
The coefficients of the (r— 1)", r and (r + 1)" terms in the
expansion of (x + 1)" are in the ratio 1: 3:5. Find n andr.
Finding (r—1) term, r? & (r +1) term of (x +1)?
Writing (x + 1)" as (1 +x)"
We know that general term of expansion (a + b)" is
Th = nC, a’~'br
For (1+x)",
Puttinga=1,b=x
Th = ne, "xt
Tra = "C, x (1)
- Coefficient of (r + 1) term ="C,
Question 11
Ex 8.2, 11 teachoo.com
Prove that the coefficient of x" in the expansion of (1+ x)*"is
twice the coefficient of x" in the expansion of (1+ x)?"~1.
We know that
General term of (a + b)" is
Tea =C, an". bt
Question 12
Ex 8.2, 12 teachoo.com
Find a positive value of m for which the coefficient of x? in the
expansion (1 +x)™is 6.
We know that
General term of expansion (a + b)" is
Test = "C, at br
General term of (1 + x)™ is
Putting n=m,a=1,b=x
Teen =C, (1)™". (x)!
=", (1). (x)!
= ™C, x"
Why Learn This With Teachoo?
Binomial Theorem provides a structured way to expand powers of a two-term expression without repeated multiplication. Students study binomial coefficients, the general term, particular coefficients, middle terms and applications to identities, comparisons and numerical evaluation. Teachoo includes Exercise 7.1, NCERT examples, miscellaneous questions and concept-wise lessons on expansion, general terms, coefficients, middle terms and proof using the binomial theorem.
Understanding the binomial expansion
For a non-negative integer n,
(a + b)ⁿ = Σ from r = 0 to n of ⁿCᵣ aⁿ⁻ʳbʳ.
Written term by term, the powers of a decrease from n to 0 while the powers of b increase from 0 to n. There are n + 1 terms before any like terms are combined. The coefficients are combinations and are symmetric because ⁿCᵣ = ⁿCₙ₋ᵣ.
For (a − b)ⁿ, substitute −b for b, producing alternating signs. Careful brackets are essential when a term itself contains a coefficient, variable power or negative sign.
General term and coefficient questions
If counting begins with the first term as T₁, the (r + 1)th term is
Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳbʳ.
This formula finds a specified term without writing the complete expansion. To find the coefficient of xᵏ, substitute the actual expressions for a and b, simplify the exponent of x in Tᵣ₊₁ and solve for the integer r. A term independent of x has total exponent zero. If the resulting r is not an integer in the range 0 ≤ r ≤ n, the requested term does not occur.
The coefficient means the numerical factor multiplying the requested power after simplification. The term itself includes both coefficient and variable part; students must distinguish them.
Middle terms
An expansion of (a + b)ⁿ contains n + 1 terms. If n is even, n + 1 is odd and there is one middle term, Tₙ/₂₊₁. If n is odd, n + 1 is even and there are two middle terms, T₍ₙ₊₁₎/₂ and T₍ₙ₊₃₎/₂. Determining the number of terms first prevents indexing errors.
Numerical and proof applications
Binomial expansions can evaluate or approximate numbers close to a convenient base, compare large powers and prove divisibility or algebraic identities. In the finite Class 11 theorem, n is a non-negative integer. Numerical approximations are obtained by choosing a small relative term and retaining enough early terms for the required accuracy.
Topics covered on Teachoo
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Exercise 7.1, NCERT examples and miscellaneous questions;
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direct binomial expansions;
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numerical evaluation using expansions;
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comparison of large expressions;
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proofs using the binomial theorem;
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definition and use of the general term;
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finding a coefficient or a specified power;
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term independent of a variable;
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one or two middle terms.
Key formulas and patterns
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(a + b)ⁿ = Σ ⁿCᵣaⁿ⁻ʳbʳ;
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Tᵣ₊₁ = ⁿCᵣaⁿ⁻ʳbʳ;
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number of terms = n + 1;
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ⁿCᵣ = ⁿCₙ₋ᵣ;
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ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ;
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sum of coefficients is obtained by setting variables equal to 1 when applicable;
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alternating sum of coefficients can often be obtained by substituting −1.
Learning outcomes
Students should be able to expand binomials for non-negative integral powers, locate a general or particular term and determine coefficients. They should identify constant and middle terms, use coefficient identities and select a convenient expansion for calculation or proof. They should understand why the term index is r + 1 while the combination index is r.
Why is this chapter important?
The theorem links algebra with combinations and prepares students for series expansions in higher mathematics. It appears in JEE questions on coefficients, divisibility, greatest terms and approximation. The general-term method also trains students to extract information without performing unnecessary expansion.
How Teachoo helps you prepare
Teachoo groups direct expansion, evaluation, comparison, proof, coefficient and middle-term problems separately. Begin by writing the first four terms using the general pattern and verify decreasing and increasing exponents. Then practise solving exponent equations using the general term.
In every answer, state Tᵣ₊₁ before substituting. This makes signs, indices and variable powers easier to audit. Use NCERT serial-order solutions for exercise completion and concept-wise practice to strengthen specific question types.
School-exam, JEE and competency preparation
School exams frequently ask for expansions, specific terms, coefficients and middle terms. JEE questions may use multiple variables, fractional expressions inside the binomial or a condition connecting coefficients. Simplify the general term completely before solving for r.
Competency questions can involve approximate calculation or error checking. Explain why a chosen base makes the second term small. If a coefficient is requested, ensure no variable remains in the reported coefficient. When comparing terms, compare absolute values if the question asks for magnitude and preserve signs if it asks for actual value.
Quick revision checklist
Expand one positive and one negative binomial; derive a general term; find coefficients of two powers; find a constant term; identify middle terms for odd and even n; use the theorem for a numerical evaluation; and prove one identity involving binomial coefficients.
Common mistakes to avoid
Do not write Tᵣ when the formula corresponds to Tᵣ₊₁. Do not forget alternating signs in (a − b)ⁿ. The number of terms is n + 1, not n. A solution for r must be an integer between 0 and n. Do not confuse the coefficient of xᵏ with the entire xᵏ term.
Deeper reasoning and concept connections
Study Binomial Theorem through comparison and justification. Place two related examples side by side, identify the decisive difference and explain why one method works in each case. Then create a new example and a deliberate non-example. This forces the definition to do real work and exposes gaps that passive reading hides.
Students should also practise reversing questions. After solving for an answer, ask what question could have produced it, whether more than one answer is possible and which extra condition would make the result unique. Reverse reasoning develops flexibility and is especially useful for missing-value, assertion–reason and error-analysis questions. The goal is to understand the network of ideas, not merely the order of a textbook solution.
How to solve unfamiliar and competency-based questions
Begin by separating facts from conclusions. Facts are given by the question or a known property; conclusions must be derived. Draw or rewrite the problem so each fact has a visible place. If several methods are possible, prefer the one with fewer assumptions and an easy final check. Record intermediate results rather than doing everything mentally.
Competency questions often change context without changing mathematics. Replace names and story details with variables, shapes, sets or data values. After solving, restore the context and check feasibility: counts should be whole where required, lengths and areas should have suitable units, probabilities should lie between 0 and 1, and constructed figures should satisfy every stated condition.
What complete mastery looks like
For Binomial Theorem, a student should be able to define the central ideas in simple language, recognise them in different representations, solve routine questions accurately and explain the method used. They should also be able to correct a flawed solution, create an example satisfying given conditions and combine two ideas from the chapter in one problem. A reliable mastery test is to solve one direct question, one application question and one reasoning question without looking at notes, then explain all three aloud or in writing.
Keep a compact error log with four labels: concept, interpretation, calculation and presentation. Reattempt each error after a gap instead of rereading the answer immediately. Improvement comes from correcting the decision that caused the mistake, not from repeating questions whose method is already known.
Additional frequently asked questions
What should a student know before starting Binomial Theorem?
Revise the definitions, number operations, diagrams or notation used at the beginning of the chapter. The prerequisite list should be short: if an earlier skill blocks progress, repair that skill with two or three focused questions and return to the chapter.
How can a student check an answer in Binomial Theorem?
Use an independent check whenever possible: substitute the result, reverse the operation, estimate its size, compare it with the figure, test a simpler case or solve using another representation. A check should examine the mathematical condition, not merely repeat the same arithmetic.
How many questions are enough for strong preparation?
There is no fixed number. Stop counting questions and track coverage: every concept, every standard method, at least one mixed problem, one competency-based problem and every previously incorrect type should be solved independently. Ten varied, analysed questions are more valuable than fifty copied solutions.
How should Teachoo solutions be used without becoming dependent on them?
Attempt the question first and mark the exact step where progress stops. Read only enough of the solution to repair that step, close it and restart the question. Finally, solve a similar problem without help. This turns a solution into feedback rather than a substitute for thinking.
Frequently asked questions
What is the general term in a binomial expansion?
The (r + 1)th term of (a + b)ⁿ is ⁿCᵣaⁿ⁻ʳbʳ.
How many terms are in (a + b)ⁿ?
There are n + 1 terms before any possible combination of like terms.
How do I find a term independent of x?
Write the general term, simplify the exponent of x, set it equal to zero and solve for a valid integer r.
When are there two middle terms?
There are two middle terms when n is odd, because the expansion then has an even number n + 1 of terms.
What does Teachoo cover in Binomial Theorem?
Teachoo covers expansions, number evaluation, comparisons, proofs, general terms, coefficients and middle terms, with NCERT answers and examples.
Use the general term as the chapter’s central tool. It turns expansion, coefficient, constant-term and middle-term problems into variations of one method.