# How to Find the Sum of n Terms of a math series?

## How to find the sum of n terms of a math series given below?

\frac{1^2}{1}+\frac{1^2+2^2 }{1+2}+\frac{1^2+2^2+3^2}{1+2+3}+....

## Solution

To find the sum of the n terms, we need to find the nth term of the math series

Let sum of the series = S_{n} and nth term = a_{n}, then

\begin{aligned} S_n&=\frac{1^2}{1}+\frac{1^2+2^2 }{1+2}+\frac{1^2+2^2+3^2}{1+2+3}+....\\ \\ \Rightarrow S_n &= \sum_{n=1}^{n} a_n \end{aligned}

All terms of this math series are rational number and numerator and denominator are another series

That is numerators are the sum of the squares and denominators are sum of the natural numbers

so n the term is

\begin{aligned} a_n &= \frac{1^2+2^2+3^2+....+n^2}{1+2+3+....+n}\\ \\ &=\frac{\frac{n(n+1)(2n+1)}{6}}{\frac{n(n+1)}{2}} \\ \\ \Rightarrow a_n&=\frac{2n+1}{3} \\ \end{aligned}

Now we get

\begin{aligned} S_n &= \sum_{n=1}^{n} \frac{2n+1}{3}\\ \\ &= \sum_{n=1}^{n} \frac{2n}{3} + \sum_{n=1}^{n} \frac{1}{3} \\ \\ &= \frac{2}{3}\sum_{n=1}^{n}n +\frac{1}{3} \sum_{n=1}^{n} 1\\ \\ &= \frac{2}{3}\times \frac{n(n+1)}{2} +\frac{1}{3} \times n\\ \\ &= \frac{n^2+n}{3} +\frac{n}{3} \\ \\ \Rightarrow S_n &= \frac{n(n+2)}{3} \\ \\ \end{aligned}

That is the sum of n terms of the series is **n(n+2)/3**