question_answer

Find the sum of n terms of the series  \[\left( 4-\frac{1}{n} \right)+\left( 4-\frac{2}{n} \right)+\left( 4-\frac{3}{n} \right)+.......\]

Answer:

In given series, \[a=\left( 4-\frac{1}{n} \right)\]

\[d=\left( 4-\frac{2}{n} \right)-\left( 4-\frac{1}{n} \right)=4-\frac{2}{n}-4+\frac{1}{n}=-\frac{1}{n}\]

\[{{S}_{n}}=\frac{n}{2}[2a+(n-1)d]\]

\[=\frac{n}{2}\left[ 2\left( 4-\frac{1}{n} \right)+(n-1)\left( -\frac{1}{n} \right) \right]\]

\[=\frac{n}{2}\left[ 8-\frac{2}{n}-\frac{(n-1)}{n} \right]\]

\[=\frac{n}{2}\left[ 7-\frac{1}{n} \right]\]

\[=\frac{n}{2}\left[ \frac{7n-1}{n} \right]\]

\[=\frac{7n-1}{2}\]