Question

If $1+2+3+......+n=45$, then find the value of ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+......+{{n}^{3}}$?
(a) ${{\left( 45 \right)}^{2}}$,
(b) ${{\left( 45 \right)}^{3}}$,
(c) ${{\left( 45 \right)}^{2}}+45$,
(d) None of these.

Answer 1

Hint: We start solving the problem by recalling the definition of the sum of first n natural numbers. We use this definition in the given $1+2+3+......+n=45$ to find the value of n. After finding the value of n, we find the value of ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+......+{{n}^{3}}$ using the fact that the sum of cubes of first n terms is $\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}$. We then make the necessary arrangements and calculations to get the required answer.

Complete step-by-step solution:
According to the problem, we have given that $1+2+3+......+n=45$ and we need to find the value of ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+......+{{n}^{3}}$.
Let us consider $1+2+3+......+n=45$ ---(1).
We know that sum of first n-natural numbers is $\dfrac{n\left( n+1 \right)}{2}$. We use this result in equation (1).
$\Rightarrow \dfrac{n\left( n+1 \right)}{2}=45$.
$\Rightarrow n\left( n+1 \right)=45\times 2$.
$\Rightarrow {{n}^{2}}+n=90$.
$\Rightarrow {{n}^{2}}+n-90=0$.
$\Rightarrow {{n}^{2}}+10n-9n-90=0$.
$\Rightarrow n\left( n+10 \right)-9\left( n+10 \right)=0$.
$\Rightarrow \left( n-9 \right)\left( n+10 \right)=0$.
$\Rightarrow n-9=0\text{ or }n+10=0$.
$\Rightarrow n=9\text{ or }n=-10$.
From equation (1), we can see that the numbers in the sum are natural numbers and all the natural numbers are positive $\left( >0 \right)$. So, we neglect $n=-10$.
So, the value of n is 9.
Now, let us find the value of ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+......+{{n}^{3}}$. We know that sum of the cubes of first n natural numbers is $\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}$.
$\Rightarrow {{1}^{3}}+{{2}^{3}}+{{3}^{3}}+......+{{n}^{3}}=\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}$.
Let us substitute 9 in place of n.
$\Rightarrow {{1}^{3}}+{{2}^{3}}+{{3}^{3}}+......+{{9}^{3}}=\dfrac{{{9}^{2}}{{\left( 9+1 \right)}^{2}}}{{{2}^{2}}}$.
$\Rightarrow {{1}^{3}}+{{2}^{3}}+{{3}^{3}}+......+{{9}^{3}}=\dfrac{{{9}^{2}}{{\left( 10 \right)}^{2}}}{{{2}^{2}}}$.
We know that ${{a}^{m}}.{{b}^{m}}={{\left( ab \right)}^{m}}$.
$\Rightarrow {{1}^{3}}+{{2}^{3}}+{{3}^{3}}+......+{{9}^{3}}=\dfrac{{{\left( 9\times 10 \right)}^{2}}}{{{2}^{2}}}$.
$\Rightarrow {{1}^{3}}+{{2}^{3}}+{{3}^{3}}+......+{{9}^{3}}=\dfrac{{{\left( 90 \right)}^{2}}}{{{2}^{2}}}$.
We know that $\dfrac{{{a}^{m}}}{{{b}^{m}}}={{\left( \dfrac{a}{b} \right)}^{m}}$.
$\Rightarrow {{1}^{3}}+{{2}^{3}}+{{3}^{3}}+......+{{9}^{3}}={{\left( \dfrac{90}{2} \right)}^{2}}$.
$\Rightarrow {{1}^{3}}+{{2}^{3}}+{{3}^{3}}+......+{{9}^{3}}={{\left( 45 \right)}^{2}}$.
We have found the value of ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+......+{{n}^{3}}$ as ${{\left( 45 \right)}^{2}}$.
$\therefore$ The value of ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+......+{{n}^{3}}$ is ${{\left( 45 \right)}^{2}}$.
The correct option for the given problem is (a).

Note: We can also see that the general sum of cubes of first n-natural numbers $\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}$ is exact square of sum of first n-natural number $\dfrac{n\left( n+1 \right)}{2}$, which gives the answer directly without making the calculations. Similarly, we can also find the sum of the squares of the given terms after finding the value of n. We should not confuse with formulas and make calculation mistakes while solving this problem.