Answer:
[tex]\boxed{2144}[/tex]
Step-by-step explanation:
The sum can be found by adding the parts:
[tex]\sum\limits_{n=1}^{32}{(4n+1)}=4\sum\limits_{n=1}^{32}{n}+\sum\limits_{n=1}^{32}{1}=4\cdot\dfrac{32\cdot 33}{2}+32\\\\= 2112+32=\boxed{2144}[/tex]
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The sum of numbers 1 to n is n(n+1)/2.
