Example 1 - Chapter 8 Class 11 Binomial Theorem

Transcript

Example, 1 Expand ﷐﷐x2 + ﷐3﷮x﷯﷯﷮4﷯ , x ≠ 0 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)4 = 4C0 a4 (b)0 + 4C1 a3 b1 + 4C2 a2 b2 + 4C3 a1b3 + 4C4 a0 b4 = ﷐4!﷮0!﷐ 4 − 0﷯!﷯ a4 × 1 + ﷐4!﷮1×(4−1)!﷯ a2 b2 + ﷐4!﷮2!﷐4 −2﷯!﷯ ab3 + ﷐4!﷮4!(4 − 4)!﷯ 1 × b4 = ﷐4!﷮1 ×4!﷯ a4 + ﷐4!﷮1 ×3!﷯ a3 b + ﷐4!﷮2!(4 − 2)!﷯ a2 b2 + ﷐4!﷮3!﷐4 −3﷯! ﷯ ab3 + ﷐4!﷮4! 0!﷯ b4 = a4 + 4a3 b + 6a2 b2 + 4 ab3 + b4 Hence, (a + b)4 = a4 + 4a3 b + 6a2 b2 + 4 ab3 + b4 Putting a = x2 and b = ﷐3﷮𝑥﷯ ﷐﷐x2 + ﷐3﷮x﷯﷯﷮4﷯ = ﷐﷐﷐𝑥﷮2﷯﷯﷮4﷯ + 4 ﷐﷐﷐𝑥﷮2﷯﷯﷮3﷯﷐﷐3﷮x﷯﷯ + 6﷐﷐﷐𝑥﷮2﷯﷯﷮2﷯ ﷐﷐﷐3﷮x﷯﷯﷮2﷯ + 4 ﷐﷐𝑥﷮2﷯﷯ ﷐﷐﷐3﷮x﷯﷯﷮3﷯+ ﷐﷐﷐3﷮x﷯﷯﷮4﷯ = 𝑥8 + 4x6 . ﷐3﷮x﷯ + 6𝑥4 . ﷐9﷮x2﷯ + 4 𝑥2 . ﷐27﷮𝑥3﷯ + ﷐81﷮𝑥4﷯ = 𝑥8 + 12 ﷐𝑥6﷮𝑥﷯ + 54 . ﷐𝑥4﷮𝑥2﷯ + 108 . ﷐𝑥2﷮𝑥3﷯ + ﷐81﷮𝑥4﷯ = 𝑥8 + 12 𝑥5 + 54 𝑥2 + 108 𝑥 + ﷐81﷮𝑥4﷯