Question

Enter your answer in the box

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Answer 1

`\boxed{2144}`

Explanation:

The sum can be found by adding the parts:

`\sum\limits_(n=1)^(32){(4n+1)}=4\sum\limits_(n=1)^(32){n}+\sum\limits_(n=1)^(32){1}=4\cdot(32\cdot 33)/(2)+32\\\\= 2112+32=\boxed{2144}`

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The sum of numbers 1 to n is n(n+1)/2.