Question

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 1

Let the given series be S =  \[\left( 4 - \frac{1}{n} \right) + \left( 4 - \frac{2}{n} \right) + \left( 4 - \frac{3}{n} \right) + . . . . . . . . . .\]

\[= \left[ 4 + 4 + 4 + . . . \right] - \left[ \frac{1}{n} + \frac{2}{n} + \frac{3}{n} + . . . \right]\]
\[ = 4\left[ 1 + 1 + 1 + . . . \right] - \frac{1}{n}\left[ 1 + 2 + 3 + . . . \right]\]
\[ = S_1 - S_2\]

\[S_1 = 4\left[ 1 + 1 + 1 + . . . \right]\]

\[a = 1, d = 0\]

\[ S_1 = 4 \times \frac{n}{2}\left[ 2 \times 1 + \left( n - 1 \right) \times 0 \right] \left( S_n = \frac{n}{2}\left( 2a + \left( n - 1 \right)d \right) \right)\]

\[ \Rightarrow S_1 = 4n\]

\[S_2 = \frac{1}{n}\left[ 1 + 2 + 3 + . . . \right]\]
\[a = 1, d = 2 - 1 = 1\]
\[ S_2 = \frac{1}{n} \times \frac{n}{2}\left[ 2 \times 1 + \left( n - 1 \right) \times 1 \right]\]
\[ = \frac{1}{2}\left[ 2 + n - 1 \right]\]
\[ = \frac{1}{2}\left[ 1 + n \right]\]

\[\text{ Thus } , S = S_1 - S_2 = 4n - \frac{1}{2}\left[ 1 + n \right]\]
\[S = \frac{8n - 1 - n}{2} = \frac{7n - 1}{2}\]

Hence, the sum of n terms of the series is \[\frac{7n - 1}{2}\]