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you wrote it wrong
\[\sum_{n=1}^{14}\] 3n+2
14 is on top, and n=1 is on the bottom.
\[\LARGE \sum_{n=1}^{14}~~(3n+2)\]
It will form an Arithmetic Progression..
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Yeah
So, it is.... 3(1)+2 3(2)+2 3(3)+2 3(4)+2 ..... 3(14)+2 3[ (1+14) × {14/2}] + 2 × 14
3(1), 3(2) 3(3) .... that I have is 3 times, the sequence of 1,2,3,4...13,14 and there are 14 TWOs
Go for it from here .
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