Recent content by Jhonny

I need to find an equivalent expression without the sigma notation of the expression: ∑{k=1,n} [k^2· 2^(-k)]

How can I solve: ∑{k=1,n} [k^2· 2^(-k)]

It seems to be that: Although: ∑{k=1,n} [k·f(k)] = ∑{k=1,n} [ ∑{j=k,n} [f(j)] ] holds, ∑{k=1,n} [ ∑{j=k,n} [j*f(j)] ] = ∑{k=1,n} [ ∑{j=k,n} ∑{i=j,n}[[f(i)]] ] does not hold.

I already tried that, but it didn't work. For example: ∑{k=1,n} [k·f(k)] = 1·f(1) + 2·f(2) + 3·f(3) + ... + n·f(n) = ∑{k=1,n} [ ∑{j=k,n} [f(j)] ] But, ∑{k=1,n} [k^2·f(k)] = 1·f(1) + 4·f(2) + 9·f(3) + ... + n^2·f(n) ≠ ∑{k=1,n} [ ∑{j=k,n} ∑{i=j,n}[[f(i)]] ]

I have found how to solve ∑{k=1,n} [k·f(k)], for example: ∑{k=1,n} [k·f(k)] = ∑{k=1,n} [ ∑{j=k,n} [f(j)] ] Dem: ∑{k=1,n} [k·f(k)] = 1·f(1) + 2·f(2) + 3·f(3) + ... + n·f(n) = f(1) + f(2) + f(2) + + f(3) + f(3) + f(3) + ... + f(n) + ... + f(n) = [f(1) + f(2) + f(3) + ... + f(n)] + + [f(2) +...

Hello, I can not find the way to solve the following equation: sum of k^2 f(k) from k=1 to n. In Particular, k^2 * (1/k)