Find the sum of 𝑛 terms of the series (4 −1/𝑛) + (4 −2/𝑛) + (4 −3/𝑛)

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Q) Find the sum of 𝑛 terms of the series (4 −1/𝑛) + (4 −2/𝑛) + (4 −3/𝑛) +…..

Ans:

Method 1:

The AP (4 −1/𝑛) + (4 −2/𝑛) + (4 −3/𝑛) +….. can be re-arranged as:

(4 + 4 + 4 + 4 + ….. up to 𝑛 terms ) – $\frac{1}{n}$ (1 + 2 + 3+ …. upto 𝑛 terms)

Therefore sum of n terms, $S_n = 4n - \frac{1}{\cancel {n}} \times \frac{\cancel{n}(n + 1)}{2}$

= $4n - \frac{(n + 1)}{2}$

= $\frac{(8n - n - 1)}{2}$

= $\frac{(7n - 1)}{2}$

Therefore, the sum of 𝑛 term is = $\bf{\frac{(7n - 1)}{2}}$ .

Method 2:

The AP is given as: $4 - \frac{1}{n}, 4 - \frac{2}{n}, 4 - \frac{3}{n},.....$

Its first term, $a = 4 - \frac{1}{n} = \frac{4n - 1}{n}$

Common difference, $d = n_2 - n_1$ $= (4-\frac{2}{n}) - (4-\frac{1}{n})$

$= \cancel{4} - \frac{2}{n} - \cancel{4} + \frac{1}{n}$

$= - \frac{1}{n}$

We know that Sum of n terms of an AP is given by:

$S_n = \frac{n}{2}(2a + (n - 1)d)$

By substituting values of ‘a’ and ‘d’, we get:

$S_n = \frac{n}{2}(2(\frac{4n - 1}{n}) + (n - 1) (\frac{-1}{n}))$

$= \frac{n}{2}(\frac{2(4n - 1)}{n}) - (\frac{n - 1}{n}))$

$= \frac{n}{2}(\frac{8n - 2 - (n - 1)}{n})$

$= \frac{ \cancel{n}}{2}(\frac{8n - 2 - n + 1)}{ \cancel{n}})$

$= \frac{n}{2}(\frac{7n-1)}{n})$ $= \frac{7n-1}{2}$

Therefore, the sum of 𝑛 term is = $\bf{\frac{7n - 1}{2}}$ .

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