- #1
m0286
- 63
- 0
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!
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!