Let S_n=11^3+1+21^3+2^3+1+2+31^3+2^3+3^3++1+2++n1^3+2^3++n^3.

Question

Let $S_n=\dfrac{1}{1^3}+\dfrac{1+2}{1^3+2^3}+\dfrac{1+2+3}{1^3+2^3+3^3}+\cdots+\dfrac{1+2+\cdots+n}{1^3+2^3+\cdots+n^3}$. If $100S_n=n$, then $n$ is equal to:

Accepted Answer

The $k$-th term is $\dfrac{k(k+1)/2}{[k(k+1)/2]^2}=\dfrac{2}{k(k+1)}=2\left(\dfrac1k-\dfrac1{k+1}\right)$.

Telescoping: $S_n=2\left(1-\dfrac{1}{n+1}\right)=\dfrac{2n}{n+1}$.

$$100S_n=n \Rightarrow \frac{200n}{n+1}=n \Rightarrow \frac{200}{n+1}=1 \Rightarrow n=199.$$

Solutions · 1