## Precalculus (6th Edition) Blitzer

Here, we have $a_1=1$ and $d=1$ The general formula to find the sum of the first n terms of a Geometric sequence is given as: $S_{n}=\dfrac{n(a_1+a_n)}{2}$ Now, $S_{n}=\dfrac{n(1+n)}{2}+1+2+3+......+n$ and $S_n=n+(n-1) +......+1$ $S_n+S_n=(n+1)+(n+1)+......+(n+1)$ $\implies S_n=\dfrac{n(n+1)}{2}$ Hence, $S_3=1+2+3+4+5=\dfrac{5(5+1)}{2}=15$