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Find the 13th partial sum of \[\sum_{i=1}^{\infty}9i-4\]
this is the series 5 , 14, 23 , 32.... which is arithmetic sum of n terms = (n/2) [2a + (n-1)d] where a = first term and d = common difference here n=13, a = 4 and d = 9
So the 3rd partial sum, just as an example, will be the first three terms added up: (9*1-4)+(9*2-4)+(9*3-4). Now notice how sums work, and see if you can figure it out... Or just read the answer and gain no insight by doing something purely mechanical with absolutely no purpose but to waste your life.
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