Answer
$1890$
Work Step by Step
$a_{n+1}-a_n=4(n+1)+1-(4n+1)=4$
As the difference between consecutive terms is constant, the given sequence is arithmetic.
The formula for the finite sum of the arithmetic sequence is
$S_{n}=\frac{n(a_{1}+a_{n})}{2}$
$a_{n}=4n+1$
$a_{1}=4(1)+1=5$
.
$a_{30}=4(30)+1=121$
$S_{30}=\frac{30(a_{1}+a_{30})}{2}=\frac{30(5+121)}{2}=1890$
