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

  1. Chapter 8 Class 11 Binomial Theorem
  2. Serial order wise
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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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