Difference method

**smmxwell** · January 27th 2010, 12:55 AM

By expanding (2r+1)^5-(2r-1)^5, derive a formula for $\sum{r^{4}}$ from 1 to n.

I've simplified the equation to 2(80r^4+40r^2+1), where (2r+1)^5-(2r-1)^5 is f(n+1)-f(n-1). I have no slightest inclination on where to continue. anyone care to pint me in the right direction?

**CaptainBlack** · January 27th 2010, 03:04 AM

Originally Posted by smmxwell

By expanding (2r+1)^5-(2r-1)^5, derive a formula for $\sum{r^{4}}$ from 1 to n.

I've simplified the equation to 2(80r^4+40r^2+1), where (2r+1)^5-(2r-1)^5 is f(n+1)-f(n-1). I have no slightest inclination on where to continue. anyone care to pint me in the right direction?

This question requires you to use the technique of telescoping sums.

Put:

$f(r)=(2r+1)^5$

then

$f(r-1)=(2r-1)^5$

So:

$\sum_1^k [160r^4+80r^2+2]=\sum_1^k [f(r)-f(r-1)]=f(k)-f(0)$

Hence:

$160 \sum_1^k r^4=f(k)-f(0)-\left[\sum_1^k 80r^2\right] - 2k$

CB

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