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

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

& {{S}_{k}}=\dfrac{\dfrac{k\left( k+1 \right)}{2}}{k} \\\ & =\dfrac{k+1}{2} \end{aligned}$$ Hence we have $${{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{aligned} & {{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{aligned}$$ This implies $${{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{aligned} & 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{aligned}$$ On comparing equation (4) and $$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{aligned} & \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{aligned}$$ Thus we get that $$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}$$.