sigma notation

[spacewater]

n
(Sigma sign) K^2 - ( K + 1) ^2
k = 0

K^2 = n(n-1)(2n-1) / 6 - ????

can someone direct me to the step to solve the second section of the problem?

 

[chrisr]

Assuming this is

n
Sum{k[sup:2bsvdywq]2[/sup:2bsvdywq]-(k+1)[sup:2bsvdywq]2[/sup:2bsvdywq]}
k=0

you have 2 options.

(1) The sum of k[sup:2bsvdywq]2[/sup:2bsvdywq] is n(n+1)(2n+1)/6 since it is irrelevant whether you start from 0 or 1.
The sum of (k+1)[sup:2bsvdywq]2[/sup:2bsvdywq] from k=0 is the sum of i[sup:2bsvdywq]2[/sup:2bsvdywq] from i=1 to n+1, where i=k+1,
so that sum is the same as the first one with n replaced by n+1.... (n+1)(n+2)(2n+3)/6.

Subtracting these sums gives the answer in terms of n.

(2) Multiply out (k+1)[sup:2bsvdywq]2[/sup:2bsvdywq] to get k[sup:2bsvdywq]2[/sup:2bsvdywq]+2k+1.
Now subtracting leaves you with the sum of -(2k+1) from 0 to n, which is easier.

Both give you the same result which you should work out.

 

[daon]

Or 3) Realize it is a telescoping sum:

\(\displaystyle \left [(0)^2-(1)^2 \right] +\left [(1)^2-(2)^2 \right] + ... + [(n-1)^2-(n)^2] + [(n)^2-(n+1)^2] = -(n+1)^2\)

 

[chrisr]

Absolutely!

If you observe daon's post, spacewater,
you see how superior it is to working out the sums independently,
when all the terms are really cancelling except for the last one.