Question 959118
Method #1

<img src = "http://i150.photobucket.com/albums/s91/jim_thompson5910/3-27-2015%203-19-23%20PM_zpsvphhytvr.png">

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Method #2


Notice how we add 3 to each term to get the next term (2+3 = 5, 5+3 = 8, etc). This points to the fact that we have an arithmetic sequence. 


So the common difference is d = 3


The first term is a(1) = 2


Use these two facts to get the nth term


a(n) = a(1) + d(n-1)
a(n) = 2 + 3(n-1)
a(n) = 2 + 3n-3
a(n) = 3n - 1


So because the nth term is 3n-1 this means that <img src="http://www.sciweavers.org/tex2img.php?eq=2%2B5%2B8%2B11%2B...%20%3D%20%5Csum_%7Bk%3D1%7D%5E%7B%5Cinfty%7D%283k-1%29&bc=White&fc=Black&im=jpg&fs=12&ff=arev&edit=0" align="center" border="0" alt="2+5+8+11+... = \sum_{k=1}^{\infty}(3k-1)" width="269" height="50" /> (the only difference is that I'm using k instead of n as the index)


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