Question

If $S_1,S_2,S_3,\dots ,Sn$ are the sums of infinite G.P.s. whose first terms are $1,2,3,\dots ,n$ and whose common ratios are $\frac{1}{2},\frac{1}{3},\frac{1}{4},...,\frac{1}{n+1}$ respectively, then ${\sum }_{i=1^n}{S}_i=$

• A. $\frac{n\left(n+3\right)}{2}$
• B. $\frac{n\left(n+4\right)}{2}$
• C. $\frac{n\left(n-3\right)}{2}$
• D. $\frac{n\left(n+1\right)}{2}$

Answer 1

The correct option is A

$\frac{n\left(n+3\right)}{2}$

Explanation for the correct option

Step 1:Solve using sum of infinite terms of GP

We know that the formula for the sum of the G.P with infinite terms is given by,

${\mathit{S}}_{\mathbf{\infty }}\mathbf{=}\frac{\mathbf{a}}{\mathbf{1}\mathbf{-}\mathit{r}}$, where $a$ is the initial term and $r$ is the common ratio.

Then the given summation can be written as,

${\sum }_{i=1^n}{S}_i=={\sum }_{i=1^n}\frac{a}{1-r}$

It is given that initial terms and common ratio of infinite G.P's are $1,2,3,\dots ,n$ and $\frac{1}{2},\frac{1}{3},\frac{1}{4},...,\frac{1}{n+1}$ respectively.

From this, we can find that

$\begin{array}{rcl}{S}_1& =& \frac{1}{1-{\displaystyle \frac{1}{2}}}\\ & =& 2\end{array}$

$S_2$ will be,

$\begin{array}{rcl}{S}_2& =& \frac{2}{1-{\displaystyle \frac{1}{3}}}\\ & =& 3\end{array}$

$S_3$ can be found as

$\begin{array}{rcl}{S}_3& =& \frac{3}{1-{\displaystyle \frac{1}{4}}}\\ & =& 4\end{array}$

Similarly, $S_n$ will be

$\begin{array}{rcl}{S}_n& =& \frac{n}{1-{\displaystyle \frac{1}{n+1}}}\\ & =& n+1\end{array}$

Step 2: Calculation for the required summation

With the help of the above values, we can write the ${\sum }_{i=1^n}{S}_i$ as

$\begin{array}{rcl}{\sum }_{i=1^n}{S}_i& =& S_1+S_2+S_3+\dots +S_n\\ & =& 2+3+4+\dots +n+1\end{array}$

We can see that on the RHS there is an A.P with an initial term $a=2$, common difference $d=1$, and $n\text{th}$ term is $n$.

Now, the formula of the sum of an A.P is given by

${\mathit{S}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{n}}{\mathbf{2}}\left(2a+\left(n-1\right)d\right)$

Then, the sum will be

$\begin{array}{rcl}{S}_n& =& \frac{n}{2}\left(2\left(2\right)+\left(n-1\right)1\right)\\ & =& \frac{n}{2}\left(3+n\right)\end{array}$

With the help of this, we get

$\begin{array}{rcl}{\sum }_{i=1^n}{S}_i& =& \frac{n\left(3+n\right)}{2}\end{array}$

Hence, the correct option is (A).