Question

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

Answer 1

if $n=\infty$ & $\left|r\right|<1$ then formula for total sum is

$S=\frac{a}{1-r}$
Then, we have
$S_1=\frac{1}{1-\frac{1}{2}}=2$
$S_2=\frac{2}{1-\frac{1}{3}}=3$
$S_3=\frac{3}{1-\frac{1}{4}}=4$
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.
.
.
$S_p=\frac{p}{1-\frac{1}{p+1}}=p+1$
Adding all sums, we get
$\sum _{n=1}^p{S}_n=2+3+4+5.........+p+1$
$=\frac{p}{2}\times (2+p+1)$
$\therefore S_1+S_2+.......+S_p=\frac{p}{2}(p+3)$