Question

Assertion :For $n\ge 4$. Let $a_n=\sum _{j=1}^n\sum _{k=j}^n\left(\begin{array}{c}n\\ k\end{array}\right)\left(\begin{array}{c}k\\ j\end{array}\right)$ and $b_n=a_n+2^{n+1}$;

$b_{n+1}=5b_n-6b_{n-1}$ for all $n\ge 2$

Reason: $a_n=5a_n-6a_n$ for all $n\ge 2$

• 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. Assertion is incorrect but Reason is correct

Answer 1

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

$\left(\begin{array}{c}n\\ k\end{array}\right)\left(\begin{array}{c}k\\ j\end{array}\right)=\frac{n!}{k!(n-k)!}.\frac{k!}{j!(k-j)!}$

$=\frac{n!}{j!(n-j)!}\frac{(n-j)!}{(n-k)!(k-j)!}=\left(\begin{array}{c}n\\ k\end{array}\right)\left(\begin{array}{cc}n-& j\\ n-& k\end{array}\right)$
Thus $a_n=\sum _{j=1}^n\left(\begin{array}{c}h\\ k\end{array}\right)\sum _{j=1}^n\left(\begin{array}{cc}n-& j\\ n-& k\end{array}\right)$

$=\sum _{j=1}^n\left(\begin{array}{c}n\\ j\end{array}\right)2^{n-j}={(2+1)}^n-2^n=3^n-2^n$

$\Rightarrow b_1=3^n{-2}^n+2\left(2^n\right)=3^n+2^n$
Now $a_{n+1}=3^{n+1}-2^{n+1}=(3+2)(3^n-2^n)-2\left(3^n\right)+3\left(2^n\right)$

$=5(3^n-2^n)-6(3^{n-1}-2^{n+1})={5a}_n-{6a}_{n-1}$
And ${5b}_n-{6b}_{n-1}=5(a_n+2^{n+1})-6(a_{n-1}+2^n)$

$={5a}_n-6b_{n-1}+2^n(10-6)$

$\Rightarrow a_{n+1}+2^{n+2}=b_{n+1}$