Q64 / 160 · AP EAMCET 2020 (21 Sep Shift 2) · Mathematics

What is the coefficient of \(\frac{y^3}{x^8}\) in \((x+y)^{-5}\), when \(\left|\frac{y}{x}\right| < 1\) ?

  • Correct answer
    A
    -35
  • B
    -30
  • C
    -25
  • D
    10

Solution

Since \((x+y)^{-5}=\frac{1}{(x+y)^5}=\frac{1}{x^5}\left(1+\frac{y}{x}\right)^{-5},\left|\frac{y}{x}\right| < 1\) \(\because\) Coefficient of \(\frac{y^3}{x^8} \operatorname{in}(x+y)^{-5}\) \(\begin{aligned} & =\text { coefficient of } \frac{y^3}{x^8} \text { in } \frac{1}{x^5}\left(1+\frac{y}{x}\right)^{-5} \\ & =\text { coefficient of } \frac{y^3}{x^3} \operatorname{in}\left(1+\frac{y}{x}\right)^{-5} \\ & =\frac{-5(-5-1)(-5-2)}{3 !}=-35 \end{aligned}\) Hence, option (a) is correct.

Hence, the correct answer is option: 'A'.

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