Question

If it is given that ${{S}_{1}}$, ${{S}_{2}}$, ${{S}_{3}}$,…${{S}_{2n}}$ are the sums of infinite geometric series whose first terms are respectively 1, 2, 3, …, 2n and common ratio are respectively $\dfrac{1}{2}$, $\dfrac{1}{3}$, …. $\dfrac{1}{2n+1}$, find the value of $S_{1}^{2}+S_{2}^{2}+S_{3}^{2}+...+S_{2n-1}^{2}$ .

Answer 1

Hint: First of all, we will see what is an infinite geometric progression and see the formula for the sum of the terms of an infinite geometric series. Then, with the given information we will find the value of terms ${{S}_{1}}$, ${{S}_{2}}$, ${{S}_{3}}$,…${{S}_{2n}}$. Once we are able to find these terms, we will the value of $S_{1}^{2}+S_{2}^{2}+S_{3}^{2}+...+S_{2n-1}^{2}$. We will use the formula for the sum of the squares of first n natural numbers given as $\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$.

Complete step-by-step solution:
A geometric series, which goes on forever and the number of terms approaches infinity is known as infinite geometric progression. This happens when the common ratio of the geometric progression is between 0 and 1.
As we go forward in the series, the fractional common ratio decreases the value of each term of the series but does not make them zero. Thus, as we go forward and the number of terms approaches infinity, the value of each term approaches zero but is never equal to zero.
The sum of all terms of an infinite geometric series is given as $\dfrac{a}{1-r}$, where a is the first term and r is the common ratio.
It is given that ${{S}_{1}}$ is the sum of all terms of infinite geometric series with the first term as 1 and common ratio as $\dfrac{1}{2}$.
Therefore, ${{S}_{1}}=\dfrac{1}{1-\dfrac{1}{2}}$
$\Rightarrow {{S}_{1}}=2$
Similarly, it is given that ${{S}_{2}}$ is the sum of all terms of infinite geometric series with the first term as 2 and common ratio as $\dfrac{1}{3}$.
Therefore, ${{S}_{2}}=\dfrac{2}{1-\dfrac{1}{3}}$
$\Rightarrow {{S}_{2}}=3$
So, if ${{S}_{2n-1}}$ is the sum of all terms of infinite geometric series with first term as 2n – 1 and common ratio as 2n.
$\begin{align}
  & \Rightarrow {{S}_{2n-1}}=\dfrac{2n-1}{1-\dfrac{1}{2n}} \\
 & \Rightarrow {{S}_{2n-1}}=2n \\
\end{align}$.
Now, we will find the value of $S_{1}^{2}+S_{2}^{2}+S_{3}^{2}+...+S_{2n-1}^{2}$ by substituting all the values.
$\Rightarrow S_{1}^{2}+S_{2}^{2}+S_{3}^{2}+...+S_{2n-1}^{2}={{2}^{2}}+{{3}^{2}}+{{4}^{2}}+...+{{\left( 2n \right)}^{2}}$
But we know that sum of squares of n natural numbers is given as $\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$.
Therefore, ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+...+{{\left( 2n \right)}^{2}}=\dfrac{2n\left( 2n+1 \right)\left( 4n+1 \right)}{6}$
$\begin{align}
  & \Rightarrow {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+...+{{\left( 2n \right)}^{2}}=\dfrac{n\left( 2n+1 \right)\left( 4n+1 \right)}{3} \\
 & \Rightarrow {{2}^{2}}+{{3}^{2}}+...+{{\left( 2n \right)}^{2}}=\dfrac{n\left( 2n+1 \right)\left( 4n+1 \right)}{3}-1 \\
\end{align}$
Hence, $S_{1}^{2}+S_{2}^{2}+S_{3}^{2}+...+S_{2n-1}^{2}=\dfrac{n\left( 2n+1 \right)\left( 4n+1 \right)}{3}-1$.

Note: The prerequisites of this problem are geometric progressions and basic number theory. Students are advised to be careful while substituting the 2n is the place of n in the formula for the sum of squares of n natural numbers. Sum of first n natural numbers is given as $\dfrac{n\left( n+1 \right)}{2}$, whereas sum of cubes of n natural numbers is given as $\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}$.