3*1^2+4*2^2+5*3^2+...............+(n+2)*n^2 is a A.P series. Find the sumtotal.

This can be expressed as follows:

As Melody noted, it is not an Arithmetic Progression, which has a constant difference between successive terms.

.

This series begins as follows: 0, 3, 16, 45, 96, 175, 288, 441, 640, 891.......etc.

Its sum can be obtained thus:Sum_(n=1)^10(n^3-n^2-n+1) = 2595......for the first 10 terms.......etc.

Thanks Guest, 

Firstly this is not an AP

3*1^2+4*2^2+5*3^2+...............+(n+2)*n^2 is a A.P series. Find the sumtotal.

I put this into Wolfram|alpha and got this result.

https://www.wolframalpha.com/input/?i=sum+of+(k%2B2)*k%5E2+from+k%3D1+to+n

I do not know how to do this without Wolfram|alpha though.

I have seen Geno314 do questions along this line but I have not learned his technique :///

Melody,

These are quite well known summations, though I can't think, offhand, exactly how to do them!

Thanks Alan :)