Question

$4+18+48+...n$ terms = ?

• A. $\frac{n(n+1)(n+2)(3n+5)}{12}$
• B. $\frac{n(n+1)(n+2)(5n+3)}{12}$
• C. $\frac{n(n+1)(n+2)(7n+1)}{12}$
• D. $\frac{n(n+1)(n+2)(9n-1)}{12}$

Answer 1

The correct option is A

$\frac{n(n+1)(n+2)(3n+5)}{12}$

Explanation for the correct option:

Step 1: Find the $n$th term of the series.

Analyze the series to get $n$th term

Given: $4+18+48+...n$

It can also be written as $1{\left(1+1\right)}^2+2{(2+1)}^2+3{\left(3+1\right)}^2+.....n{\left(n+1\right)}^2$

So the summation of the series can be written as $\sum _1^{n}n{(n+1)}^2$.

Step 2: Find the sum using basic summation formulas.

Simplify the $n$th term and use the basic summation formulas.

$\begin{array}{rcl}\sum _1^{n}n{(n+1)}^2& =& \sum _1^n{n}^3+\sum _1^{n}2·n^2+\sum _1^{n}n\\ & =& {\left(\frac{n\left(n+1\right)}{2}\right)}^2+2·\left(\frac{n\left(n+1\right)\left(2n+1\right)}{6}\right)+\frac{n\left(n+1\right)}{2}\\ & =& \frac{n\left(n+1\right)}{2}\left(\frac{n\left(n+1\right)}{2}+\frac{2\left(2n+1\right)}{3}+1\right)\\ & =& \frac{n(n+1)(n+2)(3n+5)}{12}\end{array}$

Hence, option (A) is the correct answer.