Question

If $\alpha =1+\frac{1}{2}+\frac{1}{3}+\cdots \cdots +\frac{1}{101}$ and $\beta =\sum _{r=1}^{99}\frac{r}{(102-r)(101-r)}$, then the value of $\alpha +\beta$ is

Answer 1

$\beta =\sum _{r=1}^{99}\frac{r}{(102-r)(101-r)}\Rightarrow \beta =\sum _{r=1}^{99}r[\frac{1}{(101-r)}-\frac{1}{(102-r)}]\Rightarrow \beta =[\frac{1}{100}-\frac{1}{101}]+2[\frac{1}{99}-\frac{1}{100}]+3[\frac{1}{98}-\frac{1}{99}]⋮+98[\frac{1}{3}-\frac{1}{4}]+99[\frac{1}{2}-\frac{1}{3}]\Rightarrow \beta =\frac{99}{2}-[\frac{1}{101}+\frac{1}{100}+\frac{1}{99}+\dots \dots +\frac{1}{3}]\Rightarrow [\frac{1}{101}+\frac{1}{100}+\frac{1}{99}+\dots \dots +\frac{1}{3}]=\frac{99}{2}-\beta \Rightarrow \alpha -\frac{1}{2}-1=\frac{99}{2}-\beta \therefore \alpha +\beta =51$