- Thread starter Colin67
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Is (2+3+4+....)(2+2^2+2^3+...) = 2+ 3*2 + 4*2^2 + 5*2^3 +... ?

Why are you having trouble finding the product? (n/2)(3+n)*2(2^n-1) = n(3+n)*(2^n-1).

Have you consider proving this by induction?

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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?

If we could use calculus the proof is easy.

Yes I looked at Wolfram link. As is clearly apparent I am still not sure how to tackle this problem.

Jomo I did try and expand brackets and got n2^n-n-3+3(2)^n, and no further.

Hence my request for help, especially because i think the basic way I am approaching this is wrong

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\(\displaystyle \ S \ = 2 + 3*2 + 4*2^2 + ...................... n*2^{n-2} + (n+1)*2^{n-1} \) ......................................................(1)

\(\displaystyle 2S \ = \ \ \ \ \ 2*2 + 3*2^2 + 4*2^3 + ............................ n*2^{n-1} + (n+1)*2^{n} \) ....................................(2)

Subtracting (2) from (1)

\(\displaystyle -S \ = \ \ \ \ \ 2 + \left [2 + 2^2 + 2^3 + ............................ 2^{n-1}\right ] - (n+1)*2^{n} \) ....................................(3)

continue......

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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.

Is that clear? Basically you are saying this: I do not know how to add 1+ pi + 12.34 + sqrt (17) so i will multiply 6 and 14, get 84 and be done. The thing is that 6*14 is NOT equal to the sum.