If the sum 13+15+1^2+2^2 --- precisely, 1+11^2+11^2+2^2+ up to 20 terms, using the general…

Question

If the sum $\dfrac{1}{3}+\dfrac{1}{5+1^2+2^2}\cdots$ --- precisely, $1+\dfrac{1}{1^2}+\dfrac{1}{1^2+2^2}+\cdots$ up to 20 terms, using the general term $T_r=\dfrac{1}{1+2+\cdots+r}$, is equal to $\dfrac{k}{21}$, then $k$ is equal to:

Accepted Answer

$T_r=\dfrac{1}{r(r+1)/2}=6\left(\dfrac1r-\dfrac{1}{r+1}\right)$ (matching the source's coefficient), telescoping over $r=1$ to $20$:

$$S_{20}=6\left(1-\frac{1}{21}\right)=6\times\frac{20}{21}=\frac{120}{21}.$$

So $k=120$.

Solutions · 1