Question on Summation of a equation

Find $$\sum_{x= 1/2}^{25 / 2} x^3 + 3x + 2.$$ Are there any simpler methods to solve this summation ? I will be thankful to any helps.

Counting is in the format of ${1 \over2}, 1, {3 \over2},....$

• Is the idea that $x$ takes values $\frac12, \frac32, \frac52,\ldots$, or that it takes values $\frac12, 1, \frac32, 2, \frac52,\ldots$? – MJD May 21 '13 at 16:37
• @MJD just a second guys I am going over the question again because there was no information on the counting and it is meaningless if we don't define the counting. – user78815 May 21 '13 at 16:41
• Assuming $x$ grows as $\frac12,\frac22,\frac32 \cdots$ put $y=2x,$ the summation will become $$\frac18\sum_{1\le y\le 25}y^3+\frac32\sum_{1\le y\le 25}y+\frac12\sum_{1\le y\le 25}1$$ – lab bhattacharjee May 21 '13 at 16:43
• @MJD it is in the format of 1/2, 1, 3/2, ... – user78815 May 21 '13 at 16:45

Put $\frac y2 = x$. Then the lower limit, which was $x=\frac12$, becomes $\frac y2=\frac12$, so $y=1$, and the upper limit, which was $x=\frac{25}2$, becomes $\frac y2=\frac{25}2$, so $y=25$. We now want to sum $y$ from 1 to 25:

$$\sum_{y=1}^{25}\left(\frac18y^3 + \frac32y + 2\right) \\ = \frac18\sum_{y=1}^{25} y^3 + \frac32\sum_{y=1}^{25} y + \sum_{y=1}^{25}2$$

Do you know how to evaluate the three sums separately?

(The important thing to check here it that we didn't somehow mess up the sums by changing the index from $x$ to $y$. There are 25 terms in the sum now, and there should have been 25 in the sum before. You said that is correct ($\frac12, 1, \frac32,\ldots \frac{25}2$ is 25 terms) so this substitution is correct. If the sum before had had only 13 terms ($\frac12,\frac32,\ldots \frac{25}2$ is only 13 terms) then new summation would not be the same.)

• Yes I do. Thank you very much. – user78815 May 21 '13 at 16:51
• I'm glad to help. – MJD May 21 '13 at 16:52
• @MJD This sum equals to 13250. However, I computed the summation given in this question with Wolfram Alpha it computes as 59189/8. I am not sure what kind of counting does Wolfram Alpha use ? – DreamLighter May 21 '13 at 19:18
• Yes I did have an error. What I got is 13.740,625. What was the output you obtained ? – DreamLighter May 21 '13 at 19:44
• That's what I get. – MJD May 21 '13 at 19:46