Question

What is this equal to?

Answer 1

`\sum_(n=1) ^(32) (4n + 1) = 2144`

Step-by-step 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= (n)/(2) (a + l)`

The sum of the first 32 terms is

`S_ {32} = (32)/(2) (5 + 129)`

`S_ {32} =16 * 134`

`S_ {32} =2144`

Therefore,

`\sum_(n=1) ^(32) (4n + 1) = 2144`