Question

Let \[{{S}_{k}}=\dfrac{1+2+3+...+k}{k}\]. If \[S_{1}^{2}+S_{2}^{2}+...+S_{10}^{2}=\dfrac{5}{12}A\], then determine the value of \[A\].
(a) 303
(b) 283
(c) 156
(d) 301

Answer 1

Hint: In this question, we are given that \[{{S}_{k}}=\dfrac{1+2+3+...+k}{k}\] where \[1+2+3+...+k\] is the sum of first \[k\] natural numbers. Now we know that the sum of first \[n\] natural numbers is given by \[\dfrac{n\left( n+1 \right)}{2}\]. Also we know that the sum of squares of first \[n\] natural numbers is given by \[\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\]. We will be using both these formula in order to evaluate the value of \[A\].

Complete step by step answer:
We are given that \[{{S}_{k}}=\dfrac{1+2+3+...+k}{k}\] where \[1+2+3+...+k\] is the sum of first \[k\] natural numbers.
Since we know that the sum of first \[n\] natural numbers is given by \[\dfrac{n\left( n+1 \right)}{2}\].
Therefore the sum of first \[k\] natural numbers \[1+2+3+...+k\] is given by
\[1+2+3+...+k=\dfrac{k\left( k+1 \right)}{2}...........(1)\]
Now on substituting the value of equation (1) in \[{{S}_{k}}=\dfrac{1+2+3+...+k}{k}\], we will have
\[\begin{align}
  & {{S}_{k}}=\dfrac{\dfrac{k\left( k+1 \right)}{2}}{k} \\
 & =\dfrac{k+1}{2}
\end{align}\]
Hence we have \[{{S}_{k}}=\dfrac{k+1}{2}\].
We are also given that \[S_{1}^{2}+S_{2}^{2}+...+S_{10}^{2}=\dfrac{5}{12}A\].
Now since we know that the sum of squares of first \[n\] natural numbers is given by \[\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\].
Therefore the sum of squares of first 10 terms of the sequence \[\left( {{S}_{k}} \right)\] is given by
\[S_{1}^{2}+S_{2}^{2}+...+S_{10}^{2}=\sum\limits_{k=1}^{k=10}{{{S}_{k}}^{2}}\]
Now on substituting the values \[{{S}_{k}}=\dfrac{k+1}{2}\] in the above equation, we get
\[S_{1}^{2}+S_{2}^{2}+...+S_{10}^{2}=\sum\limits_{k=1}^{k=10}{{{\left( \dfrac{k+1}{2} \right)}^{2}}}\]
On expanding the above summation, we have
\[S_{1}^{2}+S_{2}^{2}+...+S_{10}^{2}=\dfrac{{{2}^{2}}+{{3}^{2}}+...+{{11}^{2}}}{4}......(3)\]
Now since using the formula for the sum of squares of first \[n\] natural numbers is given by \[\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\].
The sum of squares of first 11 natural numbers is given by
\[\begin{align}
  & {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+...+{{11}^{2}}=\dfrac{11\times \left( 11+1 \right)\times \left( 2\times 11+1 \right)}{6} \\
 & =\dfrac{11\times 12\times 23}{6}
\end{align}\]
This implies \[{{2}^{2}}+{{3}^{2}}+...+{{11}^{2}}=\dfrac{11\times 12\times 23}{6}-1.........(3)\]
On substituting the value obtained in equation (3) in (2), we get
  \[\begin{align}
  & S_{1}^{2}+S_{2}^{2}+...+S_{10}^{2}=\dfrac{{{2}^{2}}+{{3}^{2}}+...+{{11}^{2}}}{4} \\
 & =\dfrac{1}{4}\left( \dfrac{11\times 12\times 23}{6}-1 \right).......(4)
\end{align}\]
On comparing equation (4) and \[S_{1}^{2}+S_{2}^{2}+...+S_{10}^{2}=\dfrac{5}{12}A\], we get
\[\dfrac{1}{4}\left( \dfrac{11\times 12\times 23}{6}-1 \right)=\dfrac{5}{12}A\]
On dividing the above equation by 4 , we get
\[\left( \dfrac{11\times 12\times 23}{6}-1 \right)=\dfrac{5}{3}A\]
Now on solving the above equation to find the value of \[A\], we will have
\[\begin{align}
  & \left( \dfrac{3036}{6}-1 \right)=\dfrac{5}{3}A \\
 & \Rightarrow \left( 506-1 \right)=\dfrac{5}{3}A \\
 & \Rightarrow 505=\dfrac{5}{3}A \\
 & \Rightarrow A=101\times 3 \\
 & \Rightarrow A=303
\end{align}\]
Thus we get that \[A=303\].

So, the correct answer is “Option A”.

Note: In this problem, we can determine the value of \[A\] using the given information. For that we are using the formula for determining the sum of first \[n\] natural numbers is given by \[\dfrac{n\left( n+1 \right)}{2}\]. Also we know that the sum of squares of first \[n\] natural numbers is given by \[\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\].