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
\[\dfrac{{{1}^{3}}+{{2}^{3}}+........+{{(2n)}^{3}}}{{{1}^{2}}+{{2}^{2}}+........+{{n}^{2}}}\]
\[\begin{align}
  & =\dfrac{{{\left\{ \dfrac{2n(2n+1)}{2} \right\}}^{2}}}{\dfrac{n(n+1)(2n+1)}{6}} \\
 & =\dfrac{6{{n}^{2}}{{(2n+1)}^{2}}}{n(n+1)(2n+1)} \\
 & =\dfrac{6n(2n+1)}{(n+1)} \\
 & =\dfrac{6n(n+1+n)}{(n+1)} \\
 & =6n+\dfrac{6{{n}^{2}}}{(n+1)} \\
 & =6n+\dfrac{6({{n}^{2}}-1)+6}{(n+1)} \\
 & =6n+\dfrac{6({{n}^{2}}-1)}{(n+1)}+\dfrac{6}{(n+1)} \\
\end{align}\]
We know the formula, \[{{a}^{2}}-{{b}^{2}}=(a+b)(a-b)\] . Now, using this formula in the above equation, we
Get,
\[\begin{align}
  & =6n+\dfrac{6({{n}^{2}}-1)}{(n+1)}+\dfrac{6}{(n+1)} \\
 & =6n+\dfrac{6(n+1)(n-1)}{(n+1)}+\dfrac{6}{(n+1)} \\
\end{align}\]
On solving we get,
\[\begin{align}
  & =6n+\dfrac{6(n+1)(n-1)}{(n+1)}+\dfrac{6}{(n+1)} \\
 & =6n+6(n-1)+\dfrac{6}{(n+1)} \\
\end{align}\]
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
\[\begin{align}
  & Integer+Integer+\dfrac{6}{(n+1)}=Integer \\
 & \Rightarrow \dfrac{6}{(n+1)}=Integer \\
\end{align}\]
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.