Did you use the link? \(\sum\limits_{k = 2}^n {k\cdot{2^{k - 1}}} = {2^n}(n - 1)\)So the AP is still 2,3,4...
The GP last term is (n-1), so sum is 2(2^(n-1) -1)
Is that correct?
You missed my point completely. When you multiply (2+3+4+....) and (2+2^2+2^3+...) do you get 2+ 3*2 + 4*2^2 + 5*2^3 +.... If not then multiplying (2+3+4+....) and (2+2^2+2^3+...) would not be helpful as it does not equal the original equation, 2+ 3*2 + 4*2^2 + 5*2^3 +...Jomo I did try and expand brackets and got n2^n-n-3+3(2)^n, and no further.