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4.   Find the sum to n terms of each of the series in \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+ ...

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Series =

               \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+ ...

n^{th}\, term=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}

a_1=\frac{1}{1}-\frac{1}{2}

a_2=\frac{1}{2}-\frac{1}{3}

a_3=\frac{1}{3}-\frac{1}{4}.................................

a_n=\frac{1}{n}-\frac{1}{n+1}

a_1+a_2+a_3+...................a_n=\left [ \frac{1}{1} +\frac{1}{2}+\frac{1}{3}+............\frac{1}{n}\right ]-\left [ \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.............\frac{1}{n+1} \right ]

a_1+a_2+a_3+...................a_n=\left [ \frac{1}{1} \right ]-\left [ \frac{1}{n+1} \right ]

                                         S_n=\frac{n+1-1}{n+1}

                                          S_n=\frac{n}{n+1}

Hence, the sum is 

                        S_n=\frac{n}{n+1}

Posted by

seema garhwal

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