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