Solve Summation Problem: \Sigma^{4}_{k=0} \stackrel{1}{k^{2}+1}

[Geekchick]

Homework Statement

$$\Sigma^{4}_{k=0}$$ $$\stackrel{1}{k^{2}+1}$$

Homework Equations

I would imagine it has something to do with this property

$$\Sigma^{n}_{i=1}$$ $$i^{2}$$ = $$\stackrel{n(n+1)(2n+1)}{6}$$

The Attempt at a Solution

So at first I thought I could bring $$k^{2}$$+1 to the top by,

$$(k^{2}+1)^{-1}$$

However that didn't work.

I do know that I can solve it by

$$\stackrel{1}{0^{2}+1}$$+$$\stackrel{1}{1^{2}+1}$$ and so on and so for so forth but I want to know how to apply the properties.

 

[Geekchick]

sorry I suppose the computer doen't like my problem...I thought i put it in right. If it helps I'll write it out. the problem is,

Sigma with the upper bound being 4 and the lower bound being 0 and the function is one divided by (k squared plus 1)

Hopefully the rest can be figured out...just think of it as a puzzle, lol.

 

[HallsofIvy]

So the problem is to add 1/(k²+ 1) for k= 0 to 4? That's just an arithmetic problem!

1/1+ 1/(4+1)+ 1/(9+1)+ 1/(16+1)= what?

 

[Geekchick]

yes but I wanted to see how this may apply to the formula [n(n+1)(2n+1)]/6 for any k squared or maybe a different formula I'm not aware of because what if the problem went form 0-150 there is no way anyone would want to work that out the way you did.