summation

**thereddevils** · September 2nd 2009, 06:32 AM

Find expression in terms of n for the sum of this series , to the number of terms stated , simplifynig the answer as far as possible .

$2^3+4^3+6^3+8^3+...$ to n terms .

My work :

$\sum^{n}_{r=1}r^3-\sum^{n}_{r=1} (2r-1)^3$

$=\frac{1}{4}n^2(n+1)^2-\sum^{n}_{r=1}(8r^3-12r^2+6r-1)$

$=\frac{1}{4}n^2(n+1)^2-8[\frac{1}{4}n^2(n+1)^2]+12(\frac{1}{6}n(n+1)(2n+1))-6(\frac{1}{2}n(n+1))+n$

I am not really sure how to put it into its simplest form from here .

i can only see that n is common .

**red_dog** · September 2nd 2009, 06:57 AM

$(2\cdot 1)^3+(2\cdot 2)^3+(2\cdot 3)^3+\ldots+(2n)^3=$

$=\sum_{k=1}^n(2k)^3=8\sum_{k=1}^nk^3=8\cdot\frac{n ^2(n+1)^2}{4}=2n^2(n+1)^2$

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