S=^-11n^2+n+1+^-11n^2+3n+3++^-111+(n+19)(n+20), then S is equal to:

Question

$S=	an^{-1}\dfrac{1}{n^2+n+1}+	an^{-1}\dfrac{1}{n^2+3n+3}+\cdots+	an^{-1}\dfrac{1}{1+(n+19)(n+20)}$, then $	an S$ is equal to:

Accepted Answer

Each term $\tan^{-1}\left(\dfrac{1}{1+(n+k)(n+k+1)}\right)=\tan^{-1}(n+k+1)-\tan^{-1}(n+k)$ telescopes over $k=0$ to $19$:

$$S=\tan^{-1}(n+20)-\tan^{-1}(n)=\tan^{-1}\left(\frac{20}{1+n(n+20)}\right)=\tan^{-1}\left(\frac{20}{n^2+20n+1}\right).$$

$$\tan S=\frac{20}{n^2+20n+1}.$$

Solutions · 1