Question 1205679
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Since the problem asks for the sum of only 5 terms, evaluating the five terms using the given formula and adding them is certainly one way to find the answer.<br>
That method would be impractical if the problem asked for the sum of the first 20 terms, or the first 100 terms.  For those problems you would need a more efficient method.<br>
And you can get good practice using those methods on this relatively easy problem.<br>
{{{sum(3n(2n-1),1,5)=sum(6n^2-3n,1,5)=6*sum(n^2,1,5)-3*sum(n,1,5)}}}<br>
Then use the following formulas (useful in many applications):<br>
{{{sum(n^2,1,k)=((k)(k+1)(2k+1))/6}}}
{{{sum(n,1,k)=((k)(k+1))/2}}}<br>
In this problem, those give us<br>
{{{sum(n^2,1,5)=(5*6*11)/6=55}}}
{{{sum(n,1,5)=(5*6)/2=15}}}<br>
Then the sum in the given problem is<br>
{{{6(55)-3(15)=330-45=285}}}<br>