MCQ
The sum of first $n$ positive integers is :
- A$\frac{{\text{n(n - 1)}}}{2}$
- B$\frac{{\text{n(n - 1)}}}{3}$
- C$\frac{{\text{n(n + 1)}}}{3}$
- ✓$\frac{{\text{n(n + 1)}}}{2}$
✓
Answer
Correct option: D.
$\frac{{\text{n(n + 1)}}}{2}$
$n$ positive integers are $\{1, 2, 3, 4, ... n\}$
Here $, a = 1, d = 2 - 1 = 1$ and $n = n$
$\text{s}_{\text{n}} = \frac{\text{n}}{2}(\text{2a}+(\text{n - 1})\text{d})$
$\text{s}_{\text{n}} = \frac{\text{n}}{2}(\text{2a}+(\text{nd} - \text{d})$
$=\frac{\text{n}}{2}(1 + \text{n})$
$ = \frac{\text{n(n + 1)}}{2}$
Here $, a = 1, d = 2 - 1 = 1$ and $n = n$
$\text{s}_{\text{n}} = \frac{\text{n}}{2}(\text{2a}+(\text{n - 1})\text{d})$
$\text{s}_{\text{n}} = \frac{\text{n}}{2}(\text{2a}+(\text{nd} - \text{d})$
$=\frac{\text{n}}{2}(1 + \text{n})$
$ = \frac{\text{n(n + 1)}}{2}$
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