Question

What is the value of the sum:
$\left( {{1}^{2}} \right)+\left( {{1}^{2}}+{{2}^{2}} \right)+\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}} \right)+\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}} \right)+.........\text{ upto n terms}$ ?

Answer 1

To find the summation of the above series, we will first find the ${{n}^{th}}$ term of the series. Then, we will generalize this term in ‘i’ and find its summation for, $i=1$ to $i=n$. This will give us the sum of the above series. Also, we will be using the summation formula of, $\sum\limits_{i=1}^{i=n}{i}$, $\sum\limits_{i=1}^{i=n}{{{i}^{2}}}$and $\sum\limits_{i=1}^{i=n}{{{i}^{3}}}$. We shall proceed in this manner to get the required solution to our problem.

Complete step by step answer:
We have given the sequence series as:
$\left( {{1}^{2}} \right)+\left( {{1}^{2}}+{{2}^{2}} \right)+\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}} \right)+\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}} \right)+.........\text{ upto n terms}$
Now, the ${{n}^{th}}$ term of this series can be written as the sum of the squares of the numbers counting from 1 to n. Mathematically, this can be written as follows:
$\begin{aligned} & \Rightarrow {{t}_{n}}={{1}^{2}}+{{2}^{2}}+{{3}^{2}}+........+{{n}^{2}} \\\ & \therefore {{t}_{n}}=\sum\limits_{i=1}^{i=n}{{{i}^{2}}} \\\ \end{aligned}$

Using the formula of summation of squares, we can write:
$\Rightarrow {{t}_{n}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$

This can be simplified as:
$\begin{aligned} & \Rightarrow {{t}_{n}}=\dfrac{\left( {{n}^{2}}+n \right)\left( 2n+1 \right)}{6} \\\ & \therefore {{t}_{n}}=\dfrac{{{n}^{3}}}{3}+\dfrac{{{n}^{2}}}{2}+\dfrac{n}{6} \\\ \end{aligned}$

Now, we need to find the summation of ${{t}_{n}}$ after generalizing it. Let the final summation be given by ‘S’. Then, ‘S’ is equal to:
$\Rightarrow S=\sum\limits_{j=1}^{n}{{{t}_{j}}}$
Where, ${{t}_{j}}$ is equal to: $\dfrac{{{j}^{3}}}{3}+\dfrac{{{j}^{2}}}{2}+\dfrac{j}{6}$ .

Putting the value of ${{t}_{j}}$ in the above expression for ‘S’, we get:
$\Rightarrow S=\sum\limits_{j=1}^{n}{\dfrac{{{j}^{3}}}{3}+\dfrac{{{j}^{2}}}{2}+\dfrac{j}{6}}$
On simplification, we get:
$\begin{aligned} & \Rightarrow S=\sum\limits_{j=1}^{n}{\dfrac{{{j}^{3}}}{3}+\sum\limits_{j=1}^{n}{\dfrac{{{j}^{2}}}{2}}+\sum\limits_{j=1}^{n}{\dfrac{j}{6}}} \\\ & \Rightarrow S=\dfrac{1}{3}\sum\limits_{j=1}^{n}{{{j}^{3}}}+\dfrac{1}{2}\sum\limits_{j=1}^{n}{{{j}^{2}}}+\dfrac{1}{6}\sum\limits_{j=1}^{n}{\dfrac{j}{6}} \\\ \end{aligned}$

Now, using these formulas:
$\begin{aligned} & \Rightarrow \sum\limits_{j=1}^{n}{{{j}^{3}}}={{\left[ \dfrac{n\left( n+1 \right)}{2} \right]}^{2}} \\\ & \Rightarrow \sum\limits_{j=1}^{n}{{{j}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \\\ & \Rightarrow \sum\limits_{j=1}^{n}{j}=\dfrac{n\left( n+1 \right)}{2} \\\ \end{aligned}$

Our expression for ‘S’ becomes:
$\Rightarrow S=\dfrac{1}{3}{{\left[ \dfrac{n\left( n+1 \right)}{2} \right]}^{2}}+\dfrac{1}{2}\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+\dfrac{1}{6}\dfrac{n\left( n+1 \right)}{2}$
$\Rightarrow S=\dfrac{n\left( n+1 \right)\left[ n\left( n+1 \right)+\left( 2n+1 \right)+\left( 1 \right) \right]}{12}$
$\begin{aligned} & \Rightarrow S=\dfrac{n\left( n+1 \right)\left( {{n}^{2}}+3n+2 \right)}{12} \\\ & \Rightarrow S=\dfrac{n\left( n+1 \right)\left( n+1 \right)\left( n+2 \right)}{12} \\\ & \therefore S=\dfrac{n{{\left( n+1 \right)}^{2}}\left( n+2 \right)}{12} \\\ \end{aligned}$
Therefore, the value of ‘S’ is equal to $\dfrac{n{{\left( n+1 \right)}^{2}}\left( n+2 \right)}{12}$.
Hence, the summation of the series given in our problem comes out to be $\dfrac{n{{\left( n+1 \right)}^{2}}\left( n+2 \right)}{12}$.

Note: In sequential problems like these, we should always remember the sum of sequence series of terms up to their fourth power. Also, these are some of the lengthy problems of sequence and series, so one should be careful while solving these in time bound exams as they take more time to solve.