So I have the following: $\sum_{n=1}^x (\frac76)^{n-1} *n $
When I plug it into wolfram alpha it gives me the following equation: $6^{1-x}(7^xx-6*7^x+6^{x+1}) $ I have no idea where this came from. I'm guessing it involves taking the derivative to make it a normal geometric series, but I have no idea. Any help would be appreciated.
In general, how would could I find an equation for $\sum_{n=1}^x a^n*p(n) $ where a is a rational number and p is a polynomial?
Let us take an example for illustration purposes.
Say $p(n)=b_0+b_1 n + b_2 n^2+b_3 n^3$; making $$S=\sum_{n=1}^x a^n p(n)=b_0\sum_{n=1}^x a^n+b_1\sum_{n=1}^x n a^n+b_2\sum_{n=1}^x n^2 a^n+b_3\sum_{n=1}^x n^3 a^n$$ Now, the idea is to write $$n^2=n(n-1)+n$$ $$n^3=n(n-1)(n-2)+3n(n-1)+n$$ $$n^4=n(n-1)(n-2)(n-3)+6n(n-1)(n-2)+7n(n-1)+n$$ and so on making, for the limited example, $$S=b_0\sum_{n=1}^x a^n+b_1 a\sum_{n=1}^x n a^{n-1}+b_2\left(a^2\sum_{n=1}^x n(n-1) a^{n-2}+ a\sum_{n=1}^x n a^{n-1}\right)+b_3\left(a^3\sum_{n=1}^x n(n-1)(n-2) a^{n-3}+ 3a^2\sum_{n=1}^x n(n-1) a^{n-2}+a\sum_{n=1}^x n a^{n-1}\right)$$ Now, let $$T=\sum_{n=1}^x a^n=\frac{a \left(1-a^x\right)}{1-a}$$ and consider the derivatives. Then the above expression write $$S=b_0T+b_1a T'+b_2(a^2T''+a T')+b_3(a^3T'''+3a^2T''+a T')$$ $$S=b_0T+a(b_1+b_2+b_3)T'+a^2(b_2+3b_3)T''+a^3 b_3 T'''$$
