Ex 8.1, 5 - Chapter 8 Class 11 Binomial Theorem

Transcript

Ex8.1, 5 Expand ﷐﷐x+﷐1﷮x﷯﷯﷮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 We need to find ﷐﷐𝑥 +﷐1﷮𝑥﷯﷯﷮6﷯ Putting a = x & b = ﷐1﷮𝑥﷯ ﷐﷐𝑥 +﷐1﷮𝑥﷯﷯﷮6﷯ = (x)6 + 6 (x)5 ﷐﷐1﷮𝑥﷯﷯ + 15 (x4) ﷐﷐﷐1﷮𝑥﷯﷯﷮2﷯ + 20 (x)3 ﷐﷐﷐1﷮𝑥﷯﷯﷮3﷯ + 15 (x)2 ﷐﷐﷐1﷮𝑥﷯﷯﷮4﷯ + 6(x)1 ﷐﷐﷐1﷮𝑥﷯﷯﷮5﷯ + ﷐﷐﷐1﷮𝑥﷯﷯﷮6﷯ = x6 + 6x4 + 15x2 + 20 + 15 × ﷐1﷮𝑥2﷯ + 6 ﷐1﷮𝑥4﷯ + ﷐1﷮𝑥6﷯ = x6 + 6x4 + 15x2 + 20 + ﷐15﷮𝑥2﷯ + ﷐6﷮𝑥4﷯ + ﷐1﷮𝑥6﷯