Question

Find the sum of n terms of the following series:

`(4 - 1/n) + (4 - 2/n) + (4 - 3/n) +` ....

Answer 1

Given:

Series: `(4 - 1/n) + (4 - 2/n) + (4 - 3/n) +` ... (n terms).

General k-th term: `T_k = 4 - k/n`, for k = 1, 2, ..., n.

This is an arithmetic progression (AP) with first term `a = 4 - 1/n` and common difference `d = (4 - 2/n) - (4 - 1/n) = -1/n`.

Use the sum formula for n terms of an AP: `S_n = (n/2)[2a + (n - 1)d]`.

Step-wise calculation:

1. Substitute `a = 4 − 1/n` and `d = -1/n` into the formula:

`S_n = (n/2)[2(4 - 1/n) + (n - 1)(-1/n)]`

2. Simplify inside brackets:

`2(4 - 1/n) = 8 - 2/n`,

`(n - 1)(-1/n) = -(n - 1)/n`, 

So, bracket = `8 - 2/n - (n - 1)/n = 8 - (n + 1)/n = 7 - 1/n`.

3. Therefore, `S_n = (n/2)(7 - 1/n)`

= `((7n)/2) - (1/2)`

= `(7n - 1)/2`

Sum of n terms = `(7n - 1)/2`.