Evaluate the given expression:
$1 + \left( {{1^2} + {2^2}} \right) + \left( {{1^2} + {2^2} + {3^2}} \right) + ..... = $

Hint: Here we can solve it by analyzing the problem and using summation on the general term. Try to convert the expression in the form of the sum of the square of the first n natural number. From there try to generalise the equation. Then apply these formula:

Sum of first n natural numbers, $\sum\limits_{k = 1}^n {k = \dfrac{{n\left( {n + 1} \right)}}{2}} $

Sum of squares to first n natural numbers $\sum\limits_{k = 1}^n {{k^2} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} $

Sum of cubes of first n natural numbers ${\sum\limits_{k = 1}^n {{k^3} = \left[ {\dfrac{{n\left( {n + 1} \right)}}{2}} \right]} ^2} = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4}$


Complete step-by-step answer:
Let ${s_n} = 1 + \left( {{1^2} + {2^2}} \right) + \left( {{1^2} + {2^2} + {3^2}} \right) + .....$

By considering the above series carefully we can write the last term of this series as ${a_n} = \left( {{1^2} + {2^2} + {3^2} + ..... + {n^2}} \right)$
Where n is any natural number

Since we know that the sum of square of first n natural numbers i.e. ${1^2} + {2^2} + {3^2} + ..... + {n^2} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$
\[ \Rightarrow {a_n} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} = \dfrac{{n\left( {2{n^2} + 3n + 1} \right)}}{6} = \dfrac{{2{n^3} + 3{n^2} + n}}{6} = \dfrac{{{n^3}}}{3} + \dfrac{{{n^2}}}{2} + \dfrac{n}{6}\]

Also, the required sum of the given series S_n can be obtained by doing summation of any k^th term where k varies from 1 to n. Therefore k^th term can be obtained by replacing n by k in the expression of a_n, we get
${a_k} = \dfrac{{{k^3}}}{3} + \dfrac{{{k^2}}}{2} + \dfrac{k}{6}$
$\therefore {S_n} = \sum\limits_{k = 1}^n {{a_k} = \sum\limits_{k = 1}^n {\left( {\dfrac{{{k^3}}}{3} + \dfrac{{{k^2}}}{2} + \dfrac{k}{6}} \right)} } = \dfrac{{\sum\limits_{k = 1}^n {{k^3}} }}{3} + \dfrac{{\sum\limits_{k = 1}^n {{k^2}} }}{2} + \dfrac{{\sum\limits_{k = 1}^n k }}{6}$ …… (1)

As we know that
Sum of first n natural numbers, $\sum\limits_{k = 1}^n {k = \dfrac{{n\left( {n + 1} \right)}}{2}} $
Sum of squares to first n natural numbers $\sum\limits_{k = 1}^n {{k^2} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} $
Sum of cubes of first n natural numbers ${\sum\limits_{k = 1}^n {{k^3} = \left[ {\dfrac{{n\left( {n + 1} \right)}}{2}} \right]} ^2} = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4}$

Using above formulas equation (1) becomes
\[{S_n} = \dfrac{{\sum\limits_{k = 1}^n {{k^3}} }}{3} + \dfrac{{\sum\limits_{k = 1}^n {{k^2}} }}{2} + \dfrac{{\sum\limits_{k = 1}^n k }}{6} = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{{4 \times 3}} + \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{{6 \times 2}} + \dfrac{{n\left( {n + 1} \right)}}{{2 \times 6}} = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{{12}} + \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{{12}} + \dfrac{{n\left( {n + 1} \right)}}{{12}}\]
\[{S_n} = \dfrac{{n\left( {n + 1} \right)}}{{12}}\left[ {n\left( {n + 1} \right) + 2n + 1 + 1} \right] = \dfrac{{n\left( {n + 1} \right)}}{{12}}\left[ {{n^2} + n + 2n + 2} \right] = \dfrac{{n\left( {n + 1} \right)\left( {{n^2} + 3n + 2} \right)}}{{12}}\] …… (2)

Converting $\left( {{n^2} + 3n + 2} \right)$ into its factors, we can write
\[{n^2} + 3n + 2 = {n^2} + n + 2n + 2 = n\left( {n + 1} \right) + 2\left( {n + 1} \right) = \left( {n + 1} \right)\left( {n + 2} \right)\]

Therefore equation (2) becomes
${S_n} = \dfrac{{n\left( {n + 1} \right)\left( {n + 1} \right)\left( {n + 2} \right)}}{{12}} = \dfrac{{n{{\left( {n + 1} \right)}^2} + 2\left( {n + 1} \right)}}{{12}}$ where n is any natural number

Note: These types of problems can be solved by analyzing the given problem and finding any general term and then applying summation on the general term. Then using the formula of sum of first n natural numbers, sum of squares of first natural numbers and sum of the cube of the first n natural numbers.