Question

The sum to 35 terms of the series, $\frac{\text{3}}{{\text{1}}^{\text{2}}}\text{+}\frac{\text{5}}{{\text{1}}^{\text{2}}\text{+}{\text{2}}^{\text{2}}}\text{+}\frac{\text{7}}{{\text{1}}^{\text{2}}\text{+}{\text{2}}^{\text{2}}\text{+}{\text{3}}^{\text{2}}}\text{+}..\text{,}\text{is}$

Answer 1

Given series $=\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+.....$

Since the $n^{th}$ term is $\frac{2n+1}{1^2+2^2+3^2+.....+n^2}$

$=\frac{2n+1}{\frac{n(n+1)(2n+1)}{6}}$

$=\frac{(2n+1)\cdot 6}{n(n+1)(2n+1)}$

$\frac{6}{n(n+1)}$

Now we have to first $35$ terms

$\sum _{n=1}^{35}\frac{6}{n(n+1)}=6\sum _{n=1}^{35}\frac{1}{n(n+1)}=6\sum _{n=1}^{35}(\frac{1}{n}-\frac{1}{n+1})$

For $n=N$

We have $\sum _{n=1}^N\left(\frac{6}{n(n+1)}\right)=6\sum _{n=1}^N(\frac{1}{n}-\frac{1}{n+1})$

$=6(1-\frac{1}{N+1})$

$\therefore$ For $N=35$

$\sum _{n=1}^{35}\frac{6}{n(n+1)}=6(1-\frac{1}{36})=6\times \frac{35}{36}=\frac{35}{6}$.