ANSWER
[tex] \sum_{n=1} ^{32} (4n + 1) = 2144[/tex]
EXPLANATION
The given series is
[tex] \sum_{n=1} ^{32} (4n + 1) [/tex]
The first term in this series is
[tex]a_1=4(1) + 1 = 5[/tex]
The last term is
[tex]l = 4(32) + 1 = 129[/tex]
The sum of the first n terms is
[tex]S_n= \frac{n}{2} (a + l)[/tex]
The sum of the first 32 terms is
[tex]S_ {32} = \frac{32}{2} (5 + 129)[/tex]
[tex]S_ {32} =16 \times 134[/tex]
[tex]S_ {32} =2144[/tex]
Therefore,
[tex] \sum_{n=1} ^{32} (4n + 1) = 2144[/tex]
