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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Binomial Theorem

NDA 2017 · Mathematics · Binomial Theorem

The expansion of (x-y)^n, n ≥ 5 is done in the descending powers of...

The expansion of (x−y)n,n≥5(x-y)^n, n \geq 5 is done in the descending powers of xx. If the sum of the fifth and sixth terms is zero, then xy\frac{x}{y} is equal to

Options

  1. An−56\frac{\mathrm{n}-5}{6}
  2. Bn−45\frac{\mathrm{n}-4}{5}
  3. Cn−45\frac{\mathrm{n}-4}{5}
  4. D6n−5\frac{6}{n-5}

Correct answer

B. n−45\frac{\mathrm{n}-4}{5}

Step-by-step solution

(x−y)n,n≥5(x-y)^{\mathrm{n}}, \mathrm{n} \geq 5 General term, Tr+1=ncr\mathrmxn−r(−y)r\mathrm{T}_{\mathrm{r}+1}=\mathrm{n}_{\mathrm{c}_{\mathrm{r}}} \mathrmx^{\mathrm{n}-\mathrm{r}}(-\mathrm{y})^{\mathrm{r}}. T5+T6=0⇒[nc4\mathrmxn−4(−y)4]+[nc5\mathrmxn−5(−y)5]=0⇒nc4\mathrmxn−4y4−nc5\mathrmxn−5y5=0⇒nc4\mathrmxn−4y4=nc5\mathrmxn−5y\mathrmxn−4−n+5y=nc5nc4⇒xy=n!5!(n−5)!×4!(n−4)!n!=4!(n−4)(n−5)!5×4!(n−5)!=n−45\begin{aligned} & \mathrm{T}_5+\mathrm{T}_6=0 \\ & \Rightarrow\left[\mathrm{n}_{\mathrm{c}_4} \mathrmx^{\mathrm{n}-4}(-\mathrm{y})^4\right]+\left[\mathrm{n}_{\mathrm{c}_5} \mathrmx^{\mathrm{n}-5}(-\mathrm{y})^5\right]=0 \\ & \Rightarrow \mathrm{n}_{\mathrm{c}_4} \mathrmx^{\mathrm{n}-4} \mathrm{y}^4-\mathrm{n}_{\mathrm{c}_5} \mathrmx^{\mathrm{n}-5} \mathrm{y}^5=0 \\ & \Rightarrow \mathrm{n}_{\mathrm{c}_4} \mathrmx^{\mathrm{n}-4} \mathrm{y}^4=\mathrm{n}_{\mathrm{c}_5} \mathrmx^{\mathrm{n}-5} \mathrm{y} \\ & \frac{\mathrmx^{\mathrm{n}-4-\mathrm{n}+5}}{\mathrm{y}}=\frac{\mathrm{n}_{\mathrm{c}_5}}{\mathrm{n}_{\mathrm{c}_4}} \Rightarrow \frac{\mathrm{x}}{\mathrm{y}}=\frac{\mathrm{n}!}{5!(\mathrm{n}-5)!} \times \frac{4!(\mathrm{n}-4)!}{\mathrm{n}!} \\ & =\frac{4!(\mathrm{n}-4)(\mathrm{n}-5)!}{5 \times 4!(\mathrm{n}-5)!}=\frac{\mathrm{n}-4}{5} \end{aligned}
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