Question

Consider that two series are given as $ {{S}_{n}}={{1}^{3}}+{{2}^{3}}+{{3}^{3}}+...........+{{n}^{3}} $ and $ {{T}_{n}}=1+2+3+..........+n $ , then:
1. $ {{S}_{n}}={{T}_{n}} $
2. $ {{S}_{n}}=T_{n}^{4} $
3. $ {{S}_{n}}=T_{n}^{2} $
4. $ {{S}_{n}}=T_{n}^{3} $

Answer 1

Hint: Use the formula of the sum of all the cubes of n natural number (i.e. $ {{1}^{3}}+{{2}^{3}}+{{3}^{3}}+...........+{{n}^{3}} $ ) is $ \sum\limits_{a=1}^{n}{{{a}^{3}}}={{\left\{ \dfrac{n\left( n+1 \right)}{2} \right\}}^{2}} $ and also use the formula of the sum of the n number which are in arithmetic progression , that is $ \begin{align}
  & {{T}_{n}}=\dfrac{n}{2}\left( 2(1)+(n-1)(1) \right) \\
 & \\
\end{align} $ , where , a = first term and d is the common difference, to find the sum of $ {{T}_{n}}=1+2+3+..........+n $ and try to relate their sum.

Complete step-by-step answer:
It is given in the question that $ {{S}_{n}}={{1}^{3}}+{{2}^{3}}+{{3}^{3}}+...........+{{n}^{3}} $ , then by using the formula of sum of the cube of n natural number (i.e. $ \sum\limits_{a=1}^{n}{{{a}^{3}}}={{\left\{ \dfrac{n\left( n+1 \right)}{2} \right\}}^{2}} $ ), we can write $ {{S}_{n}} $ as:
 $ \sum\limits_{a=1}^{n}{{{a}^{3}}}={{1}^{3}}+{{2}^{3}}+{{3}^{3}}+...........+{{n}^{3}}={{\left\{ \dfrac{n\left( n+1 \right)}{2} \right\}}^{2}} $
In the question it is given that
 $ {{1}^{3}}+{{2}^{3}}+{{3}^{3}}+...........+{{n}^{3}}={{S}_{n}} $
Hence, $ {{S}_{n}}={{\left\{ \dfrac{n(n+1)}{2} \right\}}^{2}}.....................(1) $
Now, $ {{T}_{n}}=1+2+3+..........+n $ , is an arithmetic progression in which the common difference is ‘1’ and the first term is ‘1’.
Hence, by using the formula of the Summation of first n term of the arithmetic progression, that is
 $ \begin{align}
  & {{T}_{n}}=\dfrac{n}{2}\left( 2a+(n-1)d \right)..........(2) \\
 & \\
\end{align} $ , where $ {{T}_{n}} $ will be the sum of n numbers which are in arithmetic progression.
Here, a = first term = 1
d = common difference = 1
Hence, by putting value of a and d in equation (2) we will get:
 $ \begin{align}
  & {{T}_{n}}=\dfrac{n}{2}\left( 2(1)+(n-1)(1) \right) \\
 & \\
\end{align} $ $ \begin{align}
  & {{T}_{n}}=\dfrac{n(n+1)}{2}.....................(3) \\
 & \\
\end{align} $
Now, by putting the value of equation (1) in equation (3), that is we will put $ {{T}_{n}} $ in place of $ \dfrac{n(n+1)}{2} $ in equation (1), we will get:
 $ {{S}_{n}}={{\left( {{T}_{n}} \right)}^{2}} $
Hence, $ {{S}_{n}}=T_{n}^{2} $
Hence, option 3 is our required answer.

Note: The above solution can also be found by directly using the formula of the sum of the n natural number (i.e. $ 1+2+3+..........+n $ ) is $ \sum\limits_{a=1}^{n}{a}=\dfrac{n(n+1)}{2} $ and put its value in sum of the cube of n natural number, (i.e. $ {{1}^{3}}+{{2}^{3}}+{{3}^{3}}+...........+{{n}^{3}} $ ) is $ \sum\limits_{a=1}^{n}{{{a}^{3}}}={{\left\{ \dfrac{n\left( n+1 \right)}{2} \right\}}^{2}} $ .