Browse Questions

# If $p$ = $1$ + $\large\frac{1}{2}$ + $\large\frac{1}{3}$+....+$\large\frac{1}{n}$, then $S$ = $\large\frac{1^2}{1^3}$ + $\frac{1^2+2^2}{1^3+2^3}$ + $\large\frac{1^2+2^2+3^2}{1^3+2^3+3^3}$+.... upto $n$ terms equal to

$(a)\;\frac{4}{3}\;p-1\qquad(b)\;\frac{4}{3}\;p+\frac{1}{n}\qquad(c)\;\frac{4}{3}\;p\qquad(d)\;\frac{4}{3}\;p-\frac{2}{3}\;\frac{n}{n+1}$

Answer : $\;(d)\;\frac{4}{3}\;p-\frac{2}{3}\;\frac{n}{n+1}$
Explanation : $\;a_{r}=\large\frac{\frac{1}{6}\;r(r+1)(2r+1)}{\large\frac{r^2(r+1)^2}{4}}$
$=\frac{2}{3}\;\large\frac{(2r+1)}{r(r+1)}$
$=\frac{2}{3}\;[\large\frac{1}{r}+\large\frac{1}{r+1}]$
$S=\sum_{r=1}^{n}\;a_{r}=\large\frac{2}{3}\;[(1+\large\frac{1}{2})+(\large\frac{1}{2}+\large\frac{1}{3})+....\;(\large\frac{1}{n}+\large\frac{1}{n+1})]$
$S=\large\frac{2}{3}\;[p+p-1+\large\frac{1}{n+1}]$
$S=\large\frac{4}{3}\;p-\large\frac{2}{3}\large\frac{n}{n+1}\;.$