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

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 $\begin{aligned} & {{T}_{n}}=\dfrac{n}{2}\left( 2(1)+(n-1)(1) \right) \\\ & \\\ \end{aligned}$ , 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
$\begin{aligned} & {{T}_{n}}=\dfrac{n}{2}\left( 2a+(n-1)d \right)..........(2) \\\ & \\\ \end{aligned}$ , 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:
$\begin{aligned} & {{T}_{n}}=\dfrac{n}{2}\left( 2(1)+(n-1)(1) \right) \\\ & \\\ \end{aligned}$ $\begin{aligned} & {{T}_{n}}=\dfrac{n(n+1)}{2}.....................(3) \\\ & \\\ \end{aligned}$
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}} .