To find $ S_{ 50 } $ we use formula

$$ \color{blue}{S_n = \frac{n}{2} \cdot \left(2a_1 + (n-1) \cdot d \right) } $$

In this example we have $ a_1 = 1 ~,~ d = 1 ~,~ n = 50 $. After substituting these values into the above equation, we obtain:

$$ \begin{aligned} S_n &= \frac{n}{2} \cdot \left(2a_1 + (n-1) \cdot d \right) \\[1 em] S_{ 50 } &= \frac{ 50 }{2} \cdot \left( 2 \cdot 1 + ( 50-1) \cdot 1 \right) \\[1 em] S_{ 50 } &= \frac{ 50 }{2} \cdot \left( 2 + 49 \cdot 1 \right) \\[1 em] S_{ 50 } &= \frac{ 50 }{2} \cdot \left( 2 + 49 \right) \\[1 em] S_{ 50 } &= \frac{ 50 }{2} \cdot 51 \\[1 em] S_{ 50 } &= 1275 \end{aligned}$$

The first few terms of this sequence are:

$$ 1, ~~~2, ~~~3, ~~~4, ~~~5 . . . $$

Arithmetic sequences