Ex 8.1,4 - Chapter 8 Class 11 Binomial Theorem

Transcript

Ex 8.1, 4 Expand the expression ﷐﷐﷐𝑥﷮3﷯+﷐1﷮𝑥﷯﷯﷮5﷯ 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)5 = = ﷐5!﷮0!﷐ 5 − 0﷯!﷯ a5 × 1 + ﷐5!﷮1!﷐ 5 − 1﷯!﷯ a4 b1 + ﷐5!﷮2!﷐ 5 − 2﷯!﷯ a3 b2 + ﷐5!﷮3!﷐ 5 − 3﷯!﷯ a2b3 + ﷐5!﷮4!﷐ 5 − 4﷯!﷯ a b4 + ﷐5!﷮5!﷐ 5 −5﷯!﷯ b5 × 1 = ﷐5!﷮0! × 5!﷯ a5 + ﷐5!﷮1! × 4!﷯ a4 b + ﷐5!﷮2! 3!﷯ a3 b2 + ﷐5!﷮3! 2!﷯ a2b3 + ﷐5!﷮4! 1!﷯ a b4 + ﷐5!﷮5! 0!﷯ b5 = ﷐5!﷮5!﷯ a5 + ﷐5 × 4!﷮4!﷯ a4 b + ﷐5 × 4 × 3!﷮2! 3!﷯ a3 b2 + ﷐5 × 4 × 3!﷮2 × 1 ×3!﷯ a3b2 + ﷐5 × 4 × 3!﷮3! ×1 ×3!﷯ a2b3 + ﷐5 × 4!﷮4!﷯ ab4 + ﷐5!﷮5! ﷯ b5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 Thus, (a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 We need to find ﷐﷐﷐𝑥﷮3﷯+﷐1﷮𝑥﷯﷯﷮5﷯ Putting a = ﷐𝑥﷮3﷯ & b = ﷐1﷮𝑥﷯ ﷐﷐﷐𝑥﷮3﷯+﷐1﷮𝑥﷯﷯﷮5﷯ = ﷐﷐﷐x﷮3﷯﷯﷮5﷯+ 5 ﷐﷐﷐x﷮3﷯﷯﷮4﷯ ﷐﷐1﷮x﷯﷯ + 10 ﷐﷐﷐x﷮3﷯﷯﷮3﷯ ﷐﷐﷐1﷮x﷯﷯﷮2﷯ + 10 ﷐﷐﷐x﷮3﷯﷯﷮2﷯ ﷐﷐﷐1﷮x﷯﷯﷮3﷯ + 5 ﷐﷐x﷮3﷯﷯ ﷐﷐﷐1﷮x﷯﷯﷮4﷯ + ﷐﷐﷐1﷮x﷯﷯﷮5﷯ = ﷐𝑥5﷮243﷯ + 5 ﷐﷐﷐𝑥﷮4﷯﷮81﷯﷯﷐﷐1﷮𝑥﷯﷯ + 10﷐﷐﷐𝑥﷮3﷯﷮27﷯﷯ ﷐﷐1﷮𝑥﷯﷯ + 10 ﷐﷐﷐𝑥﷮2﷯﷮9﷯﷯﷐﷐1﷮﷐𝑥﷮3﷯﷯﷯ + 5 ﷐﷐𝑥﷮3﷯﷯ + 5 ﷐﷐𝑥﷮3﷯﷯ ﷐﷐1﷮﷐𝑥﷮4﷯﷯﷯ +﷐﷐1﷮﷐𝑥﷮5﷯﷯﷯ = ﷐𝑥5﷮243﷯ + ﷐5﷮81﷯ 𝑥3 + ﷐10﷮27﷯ 𝑥 + ﷐10﷮9﷯ ﷐1﷮𝑥﷯ + ﷐5﷮3𝑥3﷯ + ﷐1﷮𝑥5﷯