Find the sum of series $1 + 2 + 3 + 4 + 5 + 6 + . . . + n$ .

Open in App

Solution

The given series is in AP with $a = 1, d = 1$

Sum of series is given as

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

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

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

Suggest Corrections

Similar questions

n

1^{2} - 2^{2} + 3^{2} - 4^{2} + 5^{2} - 6^{2} + . . .

\frac{1}{⌊ 3} + \frac{5}{⌊ 4} + \frac{11}{⌊ 5} + \frac{19}{⌊ 6} + . . . .

n

1 \times 2 + 2 \times 3 + 3 \times 4 + 4 \times 5 + \dots

Join BYJU'S Learning Program