Check sibling questions

Ex 8.1, 12 - Find (x + 1)6 + (x - 1)6 - Chapter 8 Binomial

Ex 8.1,12 - Chapter 8 Class 11 Binomial Theorem - Part 2
Ex 8.1,12 - Chapter 8 Class 11 Binomial Theorem - Part 3
Ex 8.1,12 - Chapter 8 Class 11 Binomial Theorem - Part 4
Ex 8.1,12 - Chapter 8 Class 11 Binomial Theorem - Part 5

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Transcript

Ex 8.1, 12 Find (x + 1)6 + (x – 1)6. Hence or otherwise evaluate (√2+1)^6 + (√2−1)^6. We know that (a + b)n = nC0 an + nC1 an – 1 b1 + nC2 an – 2 b2 + ….…. + nCn – 1 a1 bn – 1 + nCn bn Hence (a + b)6 = = 6!/(0! (6 − 0)!) a6 × 1 + 6!/(1! (6 − 1) !) a5 b + 6!/2!(6 − 2)! a4 b2 + 6!/3!(6 − 3)! a3 b3 + 6!/(4! (6 − 4) !)a2 b4 + 6!/5!(6 − 5)! ab5 + 6!/(6 ! (6 − 6) !) b6 = 6!/(1 × 6! ) a6 + 6!/(1 × 5!) a5 b + 6!/(2! × 4!) a4 b2 + 6!/(3! 3!) a3 b3 + 6!/(4! 2!) a2 b4 + 6!/(5! × 1) a b5 + 6!/(6! × 1) b6 = 6!/6! a6 + (6 ×5!)/(5! ) a5b + (6 × 5 × 4!)/(2 × 4!) a4 b2 + (6 × 5 × 4 × 3!)/(3 × 2 × 1 × 3!) a3 b3 + (6 × 5 × 4!)/(2 × 1 × 4!) a2 b4 + (6 × 5!)/(1 × 5!) ab5 + 6!/6! b6 = a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6 For (a – b)6 Replace b by (–b) in (1) (a + (–b)) 6 = a6 + 6a5 (–b) + 15a4 (– b)2 + 20a3 (– b)3 + 15a2 (– b)4 + 6a(– b)5 + (– b)6 (a – b)6 = a6 – 6a5 b + 15a4 b2 – 20a3 b3 + 15a2 b4 – 6ab5 + b6 Now, (a + b)6 + (a – b)6 = (a6 + 6a5 b + 15a4 b2 + 20a3 b3 + 15a2 b4 + 6ab5 + b6) + (a6 – 6a5 b + 15a4 b2 – 20a3 b3 + 15a2 b4 – 6ab5 + b6) = a6 + a6 + 15a4 b2 + 15a4 b2 + 15a2 b4 + 15a2 b4 + b6 + b6 + 6a5 b – 6a5b + 20a3 b3 – 20a3 b3 + 6ab5 – 6ab5 = 2a6 + 30a6 b2+ 30a2 b4 + 2b6 + 0 + 0 + 0 = 2a6 + 30a6 b2+ 30a2 b4 + 2b6 = 2(a6 + 15a4 b2 + 15a2 b4 + b6 ) Thus, (a + b)6 + (a – b)6 = 2(a6 + 15a4 b2 + 15a2 b4 + b6 ) Putting a = x and b = 1 (x + 1)6 + (x – 1)6 = 2(x6 + 15x4 (1)2 + 15x2 (1)4 + (1)6 ) = 2(x6 + 15x4 + 15x2 + 1) = 2 ("(" 2")3 + 15(" 2")2 + 15 (" 2") + 1" ) = 2 (8 + 60 + 30 + 1) = 2 ( 99) = 198 Thus, (√2 + 1)6 + (√2– 1 )6 = 198 Ex 8.1,12 Find (x + 1)6 + (x 1)6. Hence or otherwise evaluate 2 +1 6 + 2 1 6 . We know that (a + b)n = nC0 an b0 + nC1 an 1 b1 + nC2 an 2 b2 + . . + nCn 1 a1 bn 1 + nCn a0 bn Hence (a + b)6 = = 6! 0! 6 0 ! a6 1 + 6! 1 ! 6 1 ! a5 b + 6! 2! 6 2 ! a4 b2 + 6! 3 ! 6 3 ! a3 b3 + 6! 4 ! 6 4 ! a2 b4 + 6! 5 ! 6 5 ! ab5 + 6! 6 ! 6 6 ! 1 b6 = 6! 1 6! a6 + 6! 1 5! a5 b + 6! 2! 4! a4 b2 + 6! 3 ! 3! a3 b3 + 6! 4 ! 2! a2 b4 + 6! 5 ! 1 a b5 + 6! 6 ! 1 b6 = 6! 6! a6 + 6 5! 5! a5b + 6 5 4! 2 4! a4 b2 + 6 5 4 3! 3 2 1 3! a3 b3 + 6 5 4! 2 1 4! a2 b4 + 6 5! 1 5! ab5 + 6! 6! b6 = a6 + 6a5b + (3 5)a4b2 + (5 4)a3b3 + (3 5)a2b4 + 6ab5 + b6 = a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6 Thus, (a + b)6 = a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6 For (a b)6 Replace b by ( b) in (1) (a + ( b)) 6 = a6 + 6a5 ( b) + 15a4 ( b)2 + 20a3 ( b)3 + 15a2 ( b)4 + 6a( b)5 + ( b)6 (a b)6 = a6 6a5 b + 15a4 b2 20a3 b3 + 15a2 b4 6ab5 + b6 Now, (a + b)6 + (a b)6 = (a6 + 6a5 b + 15a4 b2 + 20a3 b3 + 15a2 b4 + 6ab5 + b6) + (a6 6a5 b + 15a4 b2 20a3 b3 + 15a2 b4 6ab5 + b6) = a6 + a6 + 15a4 b2 + 15a4 b2 + 15a2 b4 + 15a2 b4 + b6 + b6 + 6a5 b 6a5b + 20a3 b3 20a3 b3 + 6ab5 6ab5 = 2a6 + 30a6 b2+ 30a2 b4 + 6b6 + 0 + 0 + 0 = 2a6 + 30a6 b2+ 30a2 b4 + 6b6 = 2(a6 + 15a4 b2 + 15a2 b4 + b6 ) Thus, (a + b)6 + (a b)6 = 2(a6 + 15a4 b2 + 15a2 b4 + b6 ) Putting a = x and b = 1 (x + 1)6 + (x 1)6 = 2(x6 + 15x4 (1)2 + 15x2 (1)4 + (1)6 ) = 2(x6 + 15x4 + 15x2 + 1) Thus, (x + 1)6 + (x 1)6 = 2(x6 + 15x4 + 15x2 + 1) Putting x = 2 ( 2 + 1)6 + ( 2 1 )6 = 2 2 6 + 15 2 4 + 15 2 2 + 1 = 2 2 1 2 6 + 15 2 1 2 4 + 15 2 1 2 2 + 1 = 2 (2)3 + 15(2)2 + 15 (2) + 1 = 2 (8 + 60 + 30 + 1) = 2 ( 99) = 198 Thus, ( 2 + 1)6 + ( 2 1 )6 = 198

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.