Closed Form Evaluation of Sigma Notation with Exponential and Polynomial Terms

[m0286]

Hello
I have another question:
It says: Evaluate in closed form
n(sigma)i=1 (2^i - i^2)

(The n is above the sigma and the i=1 is below)
what i did so far is:
(sigma)2*2^i-1 (sigma)i^2
=(2(2^n -1)/2-1) - (n(n+1)(2n+1)/6)

now do I need to do something else to shorten this down... if so can someone help me with it...
THANKS!

 

[0rthodontist]

That's already pretty much simplified. If you had to do anything to that, you might take the 2 inside the numerator of the first fraction.