# Ex 8.1,12

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 8.1,12 Find (x + 1)6 + (x – 1)6. Hence or otherwise evaluate 2+16 + 2−16. 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 = 22 6 + 15 2 4 + 152 2 + 1 = 22126 + 15 2124 + 152122 + 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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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .