Question

The sum of $(n+1)$ terms of $\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+.......is$

• A. $\frac{n}{n+1}$
• B. $\frac{2n}{n+1}$
• C. $\frac{2}{n(n+1)}$
• D. $\frac{2(n+1)}{n+2}$

Answer 1

The correct option is B $\frac{2n}{n+1}$

$\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+n(n+1)}$

$\frac{1}{1}+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{n+1}$

we find

$1^{st}$ term = $\frac{2}{1}-\frac{2}{2}=\frac{1}{1}$

$2^{nd}$ term = $\frac{2}{2}-\frac{2}{3}=\frac{1}{3}$

$3^{rd}$ term = $\frac{2}{3}-\frac{2}{4}=\frac{1}{6}$

and so on

$n^m$ term $=\frac{2}{n}-\frac{2}{n+1}$

$(n+1{)}^m$ term $=\frac{2}{n+1}-\frac{2}{n+2}$

Now finding the sum of 4 terms;

$\frac{2}{1}-\frac{2}{2}+\frac{2}{2}-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+\frac{2}{4}-\frac{2}{5}$

we get, $\frac{2}{1}-\frac{2}{5}$

So finding the sum of all $n+1$ terms

we get, $\frac{2}{1}-\frac{2}{n+2}=2-\frac{2}{n+1}$

$\Rightarrow \frac{2(n+1)-2}{n+1}$

$\Rightarrow \frac{2n+2-2}{n+1}$

$\Rightarrow \frac{2n}{n+1}$

Hence, option (b) is correct.