Question
$$\frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + ..... + \frac{1}{{n\left( {n + 1} \right)}}\,{\text{equals}}$$
A.
$$\frac{1}{{n\left( {n + 1} \right)}}$$
B.
$$\frac{n}{{ {n + 1} }}$$
C.
$$\frac{2n}{{ {n + 1}}}$$
D.
$$\frac{2}{{n\left( {n + 1} \right)}}$$
Answer :
$$\frac{n}{{ {n + 1} }}$$
Solution :
Let, $$S = \frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + ..... + \frac{1}{{n\left( {n + 1} \right)}}$$
Now, $$n^{th}$$ term of above series $$ = {a_n} = \frac{1}{{n\left( {n + 1} \right)}}$$
$$\eqalign{
& \Rightarrow {a_n} = \frac{1}{{n\left( {n + 1} \right)}} = \frac{1}{n} - \frac{1}{{n + 1}}\left( {{\text{by fraction}}} \right) \cr
& {\text{Now, }}S = \sum {{a_n}} = \sum {\frac{1}{n} - \sum {\frac{1}{{n + 1}}} } \cr
& = \left( {1 + \frac{1}{2} + \frac{1}{3} + ..... + \frac{1}{n}} \right) - \left( {\frac{1}{2} + \frac{1}{3} + ..... + \frac{1}{n} + \frac{1}{{n + 1}}} \right) \cr
& = 1 + \left( {\frac{1}{2} - \frac{1}{2}} \right) + \left( {\frac{1}{3} - \frac{1}{3}} \right) + ..... + \left( {\frac{1}{n} - \frac{1}{n}} \right) - \frac{1}{{n + 1}} \cr
& = 1 - \frac{1}{{n + 1}} = \frac{{n + 1 - 1}}{{n + 1}} = \frac{n}{{n + 1}} \cr} $$