## Precalculus (6th Edition) Blitzer

$S_n= n^2$ verified
The sum of an arithmetic sequence is given by: $S_n=\dfrac{n}{2}[a_1+a_n]$ and the nth term for an arithmetic sequence is given by $a_n=a_1+(n-1) d$ We are given that $1+3+5+.....+(2n-1)=n^2$ Here, $a_1=1; d=2; a_n=2n-1$ The sum of an arithmetic sequence is given by: $S_n=\dfrac{n}{2}[a_1+a_n]$ $S_{10}=\dfrac{n}{2}[1+2n-1]$ or, $s_n=\dfrac{(n)(2n)}{2}= n^2$ Hence, the result has been proved.