# evaluate the sum

• Mar 29th 2010, 06:48 PM
tbenne3
evaluate the sum
• Mar 29th 2010, 07:08 PM
simplependulum
Notice that

$\sum_{k=50}^{300} = \sum_{k=1}^{300 }- \sum_{k=1}^{49 }$

and $\sum_{k=1}^n k^3 = \frac{ (n(n+1))^2 }{4}$
• Mar 29th 2010, 07:14 PM
tbenne3
Quote:

Originally Posted by simplependulum
Notice that

$\sum_{k=50}^{300} = \sum_{k=1}^{300 }- \sum_{k=1}^{49 }$

and $\sum_{k=1}^n k^3 = \frac{ (n(n+1))^2 }{4}$

so I substitute 300 and 49 for the n? I tried that but it says I got the wrong answer
• Mar 29th 2010, 07:48 PM
mr fantastic
Quote:

Originally Posted by tbenne3
so I substitute 300 and 49 for the n? I tried that but it says I got the wrong answer

Then show all your working so that we can see where you might have gone wrong.
• Mar 29th 2010, 07:57 PM
tbenne3
got it right nevermind