## Intermediate Algebra for College Students (7th Edition)

Hence, proved. $S_n=n^2$.
The given sequence is $1+3+5+....+(2n-1)$. This is the arithmetic series. Common difference $d=2$. First term $a=1$. Last term $l=(2n-1)$. Number of terms $n$. Sum of all terms is $\frac{n}{2}\cdot (a+l)$. Substitute all values. $=\frac{n}{2}\cdot (1+(2n-1))$ Simplify. $=\frac{n}{2}\cdot (1+2n-1)$ $=\frac{n}{2}\cdot (2n)$ $=n^2$.