Question

The sum of all positive integers n for which $\dfrac{{{1}^{3}}+{{2}^{3}}+........+{{(2n)}^{3}}}{{{1}^{2}}+{{2}^{2}}+........+{{n}^{2}}}$ is also an integer is?
A.8
B.9
C.15
D.Infinite

Answer 1

Hint: We know the formula of summation of $({{1}^{3}}+{{2}^{3}}+........+{{n}^{3}})$ and $({{1}^{2}}+{{2}^{2}}+........+{{n}^{2}})$ are
{{\left\\{ \dfrac{n(n+1)}{2} \right\\}}^{2}} and $\dfrac{n(n+1)(2n+1)}{6}$ . Apply these formulas for $\dfrac{{{1}^{3}}+{{2}^{3}}+........+{{(2n)}^{3}}}{{{1}^{2}}+{{2}^{2}}+........+{{n}^{2}}}$ and then we will get $\dfrac{6n(2n+1)}{(n+1)}$ . Now, transform $\dfrac{6n(2n+1)}{(n+1)}$ as $6{{n}^{2}}+6(n-1)+\dfrac{6}{(n+1)}$ . Now, solve for
$\dfrac{6}{n+1}=Integer$ and get the values of n.

Complete step-by-step answer:
According to the question, we have to find the sum of all positive integers n for which $\dfrac{{{1}^{3}}+{{2}^{3}}+........+{{(2n)}^{3}}}{{{1}^{2}}+{{2}^{2}}+........+{{n}^{2}}}$ is also an integer. So, first of all, we need to find the summation of $({{1}^{3}}+{{2}^{3}}+........+{{n}^{3}})$ and $({{1}^{2}}+{{2}^{2}}+........+{{n}^{2}})$ .
We know the formula of summation of,
({{1}^{3}}+{{2}^{3}}+........+{{n}^{3}})={{\left\\{ \dfrac{n(n+1)}{2} \right\\}}^{2}} ……………………………….(1)
$({{1}^{2}}+{{2}^{2}}+........+{{n}^{2}})=\dfrac{n(n+1)(2n+1)}{6}$ ………………………….(2)
In the question, we have
$\dfrac{{{1}^{3}}+{{2}^{3}}+........+{{(2n)}^{3}}}{{{1}^{2}}+{{2}^{2}}+........+{{n}^{2}}}$ …………………………(3)
Replacing n by 2n in the equation (1) because we have the summation of the given series till 2n terms.
\\{{{1}^{3}}+{{2}^{3}}+........+{{(2n)}^{3}}\\}={{\left\\{ \dfrac{n(n+1)}{2} \right\\}}^{2}} ……………………………..(4)
Now, from the equation (2), equation (3), and equation (4), we get
$\dfrac{{{1}^{3}}+{{2}^{3}}+........+{{(2n)}^{3}}}{{{1}^{2}}+{{2}^{2}}+........+{{n}^{2}}}$

& =\dfrac{{{\left\\{ \dfrac{2n(2n+1)}{2} \right\\}}^{2}}}{\dfrac{n(n+1)(2n+1)}{6}} \\\ & =\dfrac{6{{n}^{2}}{{(2n+1)}^{2}}}{n(n+1)(2n+1)} \\\ & =\dfrac{6n(2n+1)}{(n+1)} \\\ & =\dfrac{6n(n+1+n)}{(n+1)} \\\ & =6n+\dfrac{6{{n}^{2}}}{(n+1)} \\\ & =6n+\dfrac{6({{n}^{2}}-1)+6}{(n+1)} \\\ & =6n+\dfrac{6({{n}^{2}}-1)}{(n+1)}+\dfrac{6}{(n+1)} \\\ \end{aligned}$$ We know the formula, $${{a}^{2}}-{{b}^{2}}=(a+b)(a-b)$$ . Now, using this formula in the above equation, we Get, $$\begin{aligned} & =6n+\dfrac{6({{n}^{2}}-1)}{(n+1)}+\dfrac{6}{(n+1)} \\\ & =6n+\dfrac{6(n+1)(n-1)}{(n+1)}+\dfrac{6}{(n+1)} \\\ \end{aligned}$$ On solving we get, $$\begin{aligned} & =6n+\dfrac{6(n+1)(n-1)}{(n+1)}+\dfrac{6}{(n+1)} \\\ & =6n+6(n-1)+\dfrac{6}{(n+1)} \\\ \end{aligned}$$ We have to find the sum of all positive integers n for which $$6n+6(n-1)+\dfrac{6}{(n+1)}$$ is also an integer. $$6n+6(n-1)+\dfrac{6}{(n+1)}=Integer$$ …………………(5) It is also given in the question that n is the positive integers. Therefore, 6n and 6(n-1) are also integers. Solving equation (5), we get $$\begin{aligned} & Integer+Integer+\dfrac{6}{(n+1)}=Integer \\\ & \Rightarrow \dfrac{6}{(n+1)}=Integer \\\ \end{aligned}$$ By hit and trial, we can see that n can be 1, 2, and 5. So, Summation of all positive values of n = 1+2+5 = 8. Hence, the correct option is (A). Note: We know the formula that the summation of, $$({{1}^{3}}+{{2}^{3}}+........+{{n}^{3}})={{\left\\{ \dfrac{n(n+1)}{2} \right\\}}^{2}}$$ $$({{1}^{2}}+{{2}^{2}}+........+{{n}^{2}})=\dfrac{n(n+1)(2n+1)}{6}$$ One might use these formulas directly in $$\dfrac{{{1}^{3}}+{{2}^{3}}+........+{{(2n)}^{3}}}{{{1}^{2}}+{{2}^{2}}+........+{{n}^{2}}}$$ and write it as $$\dfrac{{{\left\\{ \dfrac{n(n+1)}{2} \right\\}}^{2}}}{\dfrac{n(n+1)(2n+1)}{6}}$$ which is wrong. Because here, the summation of the series $$\\{{{1}^{3}}+{{2}^{3}}+........+{{(2n)}^{3}}\\}$$ is up to 2n terms. So, we need the summation of the series till 2n terms.