# Ex 8.2,10 - Chapter 8 Class 11 Binomial Theorem

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex8.2, 10 The coefficients of the (r 1)th, rth and (r + 1)th terms in the expansion of (x + 1)n are in the ratio 1 : 3 : 5. Find n and r. Finding (r 1)th term , rth & (r + 1)th term of (x + 1)n Writing (x + 1)n as (1 + x)n We know that general term of expansion (a + b)n is Tr+1 = nCr an rbr For (1 + x)n , Putting a = 1 , b = a Tr+1 = nCr 1n r xr Tr+1 = nCr xr Coefficient of (r + 1)th term = nCr For rth term of (1 + x)n Replacing r with r 1 in (1) Tr 1 + 1 = nCr 1 xr 1 Tr = nCr 1 xr 1 Coefficient of (r)th term = nCr 1 For (r 1)th term of (1 + x)n Replacing r with r 2 in (1) Tr 2 + 1 = nCr 2 xr 2 Tr 1 = nCr 2 xr 2 Coefficient of (r 1)th term = nCr-2 Since the coefficient of (r 1)th, rth and (r + 1)th terms are in ratio 1 : 3 : 5 ( 1) = 1 3 2 1 = 1 3 ! 2 ![ 2) ! ! 1 ! ( 1) ! = 1 3 ! 2 ![ 2) ! 1 ![ 1) ! ! = 1 3 ! 1 2 ![ 1) ! ! 2 ![ 2) ! = 1 3 1 1 ! ( 2 )! = 1 3 1 + 1 ! ( + 2)! = 1 3 1 + 1 ! ( + 2)( +2 1)! = 1 3 1 + 1 ! ( + 2)( +1)! = 1 3 1 ( + 2) = 1 3 3(r 1) = n r + 2 3r 3 = n + 2 r n + 2 r 3r + 3 = 0 n 4r + 5 = 0 Also ( + 1) = 3 5 ! 1 ![ 1) ! ! ! ! = 3 5 ! 1 !( +1)! ! ! ! = 3 5 ! 1 ! ! ! 1 ! ( +1)! = 3 5 ! ( +1)! = 3 5 ( )! ( +1) ( )! = 3 5 + 1 = 3 5 5r = 3 (n r + 1) 5r = 3n 3r + 3 0 = 3n 3r + 3 5r 0 = 3n 8r + 3 3n 8r + 3 = 0 Now our equations are n 4r + 5 = 0 (1) & 3n 8r + 3 = 0 (2) From (1) n 4r + 5 = 0 n = 4r 5 Putting n = 4r 5 in (2) 3n 8r + 3 = 0 3(4r 5) 8r + 3 = 0 12r 15 8r + 3 = 0 12r 8r 15 + 3 = 0 4r 12 = 0 4r = 12 r = 12 4 r = 3 Putting r = 3 in n = 4r 5 n = 4(3) 5 n = 12 5 n = 7 Hence n = 7 & r = 3

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.