Mathematics - Sequence & Series Question with Solution | TestHub

$T_r=r(n-r+1)$$T_r=nr-r^2+r$

$$\begin{array}{rl}{S}_n& =\sum _{r=1}^n{T}_r=n\sum _{r=1}^{n}r-\sum _{r=1}^n{r}^2+\sum r\\ & =\frac{n\times n(n+1)}{2}-\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\\ & =\frac{n(n+1)}{2}\left[n-\frac{(2n+1)}{3}+1\right]\\ & =\frac{n(n+1)}{2}\left[\frac{3n-2n-1+3}{3}\right]\\ & =\frac{n(n+1)(n+2)}{6}\end{array}$$

So, $S_n=\frac{n(n+1)(n+2)}{6}$.

For $S_{25}$, we have:

$$S_{25}=\frac{25\times 26\times 27}{6}=25\times 13\times 9$$

Given $S_{25}=325\lambda$, we can find $\lambda$:

$$325\lambda =25\times 13\times 9$$

$$325\lambda =325\times 9$$

$$\lambda =9$$