Question

Assertion :If $x{=}^n{C}_{n-1}{+}^{n+1}{C}_{n-1}{+}^{n+2}{C}_{n-1}+...{+}^{2n}{C}_{n-1}$ then $\frac{x+1}{2n+1}$ is integer Reason: ${}^n{C}_r{+}^n{C}_{r-1}{=}^{n+1}{C}_{r}and{}^n{C}_r$ is divisible by n if n and r are co-prime

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

$x={}^n{C}_{n-1}+{}^{n+1}{C}_{n-1}+...{}^{2n}{C}_{n-1}$
$=[{}^n{C}_n+{}^n{C}_{n-1}+{}^{n+1}{C}_{n-1}+...{}^{2n}{C}_{n-1}]-{}^n{C}_n$
$=[{}^{n+1}{C}_n+{}^{n+1}{C}_{n-1}+...{}^{2n}{C}_{n-1}]-1$
:
:
$={}^{2n}{C}_n+{}^{2n}{C}_{n-1}-1$
$={}^{2n+1}{C}_n-1$
$=x$
Hence
$x+1=\frac{2n+1!}{(n+1)!.n!}$
$\frac{x+1}{2n+1}=\frac{2n!}{n!(n+1)!}$
$=\frac{2n!}{n!.n!}.\frac{1}{n+1}$
$={}^{2n}{C}_n.\frac{1}{n+1}$
Now
${}^{2n}{C}_n$ will always be divisible by $n+1$.
Hence
$\frac{x+1}{2n+1}$ will always be an integer.