Question

The $n^{th}$ term of the series $\frac{2}{1!}+\frac{7}{2!}+\frac{15}{3!}+\frac{26}{4!}+......$ is

• A. $\frac{n\left(3n-1\right)}{2\left(n\right)!}$
• B. $\frac{n\left(3n+1\right)}{2\left(n\right)!}$
• C. $\frac{n\left(3n\right)}{2\left(n\right)!}$
• D. None of these

Answer 1

The correct option is B

$\frac{n\left(3n+1\right)}{2\left(n\right)!}$

Find the given term for the series

Given series is $\frac{2}{1!}+\frac{7}{2!}+\frac{15}{3!}+\frac{26}{4!}+......$

Here the denominator terms is $1!,2!,3!,...$

So the $n^{th}$ denominator is $n!$

The numerator is $2,7,15,26$

If $N_1$ is the numerator of $a_1$

Then

$$\begin{gathered} N_2=2+5=7 \\ {N}_3=2+5+8=15 \\ {N}_4=2+5+8+11=7 \end{gathered}$$

So,

$$\begin{gathered} N_n=\frac{n}{2}\left(2a+d\left(n-1\right)\right) \\ =\frac{n}{2}\left(4+3\left(n-1\right)\right) \\ =\frac{n}{2}\left(3n+1\right) \end{gathered}$$

Therefore $n^{th}$ term is

$a_n=\frac{n\left(3n+1\right)}{2\left(n\right)!}$

Hence, the correct option is $\left(\mathrm{B}\right)$