Question

If $S_1,S_2,S_3\cdots S_p$ are the sums of infinite geometric series whose first terms are $1,2,3,...p$ and whose common
ratios are $\frac{1}{2},\frac{1}{3},\frac{1}{4},\cdots \frac{1}{p+1}respectively,$ then $S_1+S_2+S_3+\cdots +S_p=$ ___.

• A. $\frac{p(p+1)}{2}$
• B. $\frac{p(p+3)}{2}$
• C. p
• D. None of these

Answer 1

The correct option is B

$\frac{p(p+3)}{2}$

$S_1=\frac{a}{1-r}=\frac{1}{1-\frac{1}{2}}=2,$
$S_2=\frac{2}{1-\frac{1}{3}}=\frac{2}{\frac{2}{3}}=3,S_3=\frac{3}{1-\frac{1}{4}}=\frac{3}{\frac{3}{4}}=4$
.
.
.
$S_P=\frac{p}{1-\frac{1}{p+1}}=\frac{p}{\frac{p}{p+1}}=p+1$

$\therefore S_1+S_2+S_3+\cdots +S_p=2+3+4+........+p+1$
$=\frac{p}{2}(2\times 2+(p-1\left)1\right)=\frac{p(p+3)}{2}(2,3,4,...,(p+1\left)\text{are in AP}\right)$
(Sum of $n$ terms in a A.P = $\frac{n}{2}(2a+(n-1\left)d\right)$