The sum of the first 9 terms of the series whose n-th term is T_n=(1+2++nn)^2 is:

Question

The sum of the first 9 terms of the series whose $n$-th term is $T_n=\left(\dfrac{1+2+\cdots+n}{n}\right)^2$ is:

Accepted Answer

$T_n=\left(\dfrac{n(n+1)/2}{n}\right)^2=\dfrac{(n+1)^2}{4}$.

$$S_9=\frac14\sum_{n=1}^{9}(n+1)^2=\frac14\big(2^2+3^2+\cdots+10^2\big)=\frac14\left(\sum_{k=1}^{10}k^2-1\right)=\frac14(385-1)=96.$$

Solutions · 1