Question

What is this equal to?

Answer 1

ANSWER

$\sum_{n=1} ^{32} (4n + 1) = 2144$

EXPLANATION

The given series is

$\sum_{n=1} ^{32} (4n + 1)$

The first term in this series is

$a_1=4(1) + 1 = 5$

The last term is

$l = 4(32) + 1 = 129$

The sum of the first n terms is

$S_n= \frac{n}{2} (a + l)$

The sum of the first 32 terms is

$S_ {32} = \frac{32}{2} (5 + 129)$

$S_ {32} =16 \times 134$

$S_ {32} =2144$

Therefore,

$\sum_{n=1} ^{32} (4n + 1) = 2144$