Question

If $f\left(r\right)=1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{r}$ and $f\left(0\right)=0$, then value of $\sum _{r=1}^n(2r+1)f\left(r\right)$ is

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

Answer 1

The correct option is B $(n+1{)}^{2}f(n+1)-\frac{n^2+3n+2}{2}$

Since $\sum _{r=1}^n(2r+1)f\left(r\right)=\sum _{r=1}^n(r^2+2r+1-r^2)f\left(r\right)=\sum _{r=1}^n\left\{\right(r+1{)}^2-r^2\}f\left(r\right)$
$=\sum _{r=1}^n\left\{\right(r+1{)}^{2}f\left(r\right)-(r+1{)}^{2}f(r+1)+(r+1{)}^{2}f(r+1)-r^{2}f\left(r\right)\}$
$=\sum _{r=1}^n(r+1{)}^2\left\{f\right(r)-f(r+1\left)\right\}+\sum _{r=1}^n\left\{\right(r+1{)}^{2}f(r+1)-r^{2}f\left(r\right)\}$
$=-\sum _{r=1}^n\frac{(r+1{)}^2}{(r+1)}+\sum _{r=1}^{n-1}(r+1{)}^{2}f(r+1)+(n+1{)}^{2}f(n+1)-\sum _{r=1}^n{r}^{2}f\left(r\right)$
$=-\sum _{r=1}^n(r+1)+\left\{2^{2}f\right(2)+3^{2}f(3)+....+n^{2}f(n\left)\right\}+(n+1{)}^{2}f(n+1)-\left\{1^{2}f\right(1)+2^{2}f(2)+3^{2}f(3)+.....+n^{2}f(n\left)\right\}$
$=-\sum _{r=1}^{n}r-\sum _{r=1}^{n}1+\left(\right(n+1{)}^{2}f(n+1)-1^{2}f\left(1\right))=-\frac{n(n+1)}{2}-n+(n+1{)}^{2}f(n+1)-f(1)$
$=(n+1{)}^{2}f(n+1)-\frac{n(n+3)}{2}-1=(n+1{)}^{2}f(n+1)-\frac{(n^2+3n+2)}{2}$