I'm having a hard time trying to find a closed form for $(2^0)(1)(2) + (2^1)(2)(3) + (2^2)(3)(4) + ... + (2^{n-2})(n-1)(n)$. I know that $1\cdot2+2\cdot3+3\cdot4+...+n(n+1) = n(n+1)(n+2)/3$, but I don't know how to incorporate that into my solution.
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Are you familiar with $\sum_{k=0}^{N-1} x^k = (x^N-1)/(x-1) ?$ Are you familiar with $d/dx,x^k = k,x^{k-1}$? If so with these two basic tools and a little ingenuity of where you put the $x$'s, you can construct a proof for $\sum_{k=0}^{n-2} x^k,(k+1),(k+2) $ and then put $x=2$ afterwards. – user321120 Oct 16 '18 at 16:49
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See https://math.stackexchange.com/questions/593996/how-to-prove-sum-n-0-infty-fracn22n-6 – lab bhattacharjee Oct 16 '18 at 17:08
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We have $$f(x)=\sum_{k=0}^{n}x^k={1-x^n\over 1-x}$$for $x\ne 1$ and $f(x)=n+1$ for $x=1$. Therefore $$f''(x){=\sum_{k=2}^{n}k(k-1)x^{k-2}\\=\sum_{k=0}^{n-2}(k+2)(k+1)x^k\\={n(n-1)x^{n-2}\over x-1}-{2\over (x-1)^3}\Bigg((n-1)x^n-nx^{n-1}+1\Bigg)}$$and by substituting $x=2$ we finally conclude that $$f''(2){=\sum_{k=2}^{n}k(k-1)2^{k-2}\\=n(n-1)2^{n-2}-(n-1)2^{n+1}+n\cdot2^{n}-2\\=(n^2-5n+8)2^{n-2}+2}$$
Mostafa Ayaz
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