Find the sum to n terms of the series 12 + (12 + 22) + (12 + 22 + 32) + …

Given series is 1² + (1² + 2²) + (1² + 2² + 3² ) + …

Finding the n^th term, we have

a_n = (1² + 2² + 3² +…….+ n²)

= \( \dfrac{n(n+1)(n+1)}{6}\)

= \( \dfrac{n(2n^2+3n+1)}{6}\)

= \( \dfrac{2n^2+3n^2+n}{6}\)

= \( \dfrac{1}{3}n^3+\dfrac{1}{2}n^2+\dfrac{1}{6}n\)

The sum of n terms of a series is given by the equation 

\( S_n= \displaystyle\sum_{k=1}^{n}a_k\)

The sum of n terms of the given series is

\( S_n= \displaystyle\sum_{k=1}^{n}= (\dfrac{1}{3}k^3+\dfrac{1}{2}k^2+\dfrac{1}{6}k)\)

\( \dfrac{1}{3}\displaystyle\sum_{k=1}^{n}=k^3+\dfrac{1}{2}\displaystyle\sum_{k=1}^{n}k^2+\dfrac{1}{6}\displaystyle\sum_{k=1}^{n}k\)

\( \dfrac{1}{3}\dfrac{n^2(n+1)^2}{(2)^2}+\dfrac{1}{2}\times \dfrac{n(n+1)(2n+1)}{6}\)

\(+ \dfrac{1}{6}\times \dfrac{n(n+1)}{2}\)

\( \dfrac{n(n+1)}{6}[\dfrac{n(n+1)}{2}+\dfrac{(2n+1)}{2}+\dfrac{1}{2}]\)

\( \dfrac{n(n+1)}{6}[\dfrac{n^2+n+2n+2}{2}]\)

\( \dfrac{n(n+1)}{6}[\dfrac{(n+1)+2(n+1)}{2}]\)

\( \dfrac{n(n+1)}{6}[\dfrac{(n+1)+(n+2)}{12}]\)

= \( \dfrac{n(n+1)(n+2)}{12}\)