We have to find the sum to nth term in the following series:
$$1-\frac{2}{2}+\frac{3}{2^2}-\frac{4}{2^3}+$$ up to nth term.
I tried using the common method of successive differences. It lead me to an answer that was:
$$\frac{3}{2}S=\frac{(-2)^n2+4}{3}+\frac{2-3n}{(-2)^n}$$
Where S is the sum of the series.
I'm not sure if this is the answer. Could anyone help me out by checking this answer and recommend a better, not so sophisticated method.
On
We are to find the value of $$\sum_{i=1}^{n}{i\over (-2)^{i-1}}$$define $$f(x)=\sum_{i=1}^{n}x^i=x{1-x^{n}\over 1-x}$$therefore$$f'(x)=\sum_{i=1}^{n}ix^{i-1}={(1-(n+1)x^n)(1-x)+x-x^{n+1}\over (1-x)^2}={1-(n+1)x^n+nx^{n+1}\over (1-x)^2}$$by setting $x=-{1\over 2}$ we finally obtain$$\sum_{i=1}^{n} i\left(-{1\over 2}\right)^{i-1}={4\over 9}\left[1+(3n+2)\left(-{1\over 2}\right)^{n+1}\right]$$
A possible way:
We want to compute
\begin{align}S_n &= \sum_{i=0}^{n-1} \frac{i+1}{(-2)^i}\\&=\sum_{i=0}^{n-1} (i+1)(-0.5)^{i}\\&= \sum_{i=0}^{n-1} \frac{d}{dx}x^{i+1}|_{x=-0.5} \\ &= \frac{d}{dx}\left(\sum_{i=0}^{n-1} x^{i+1} \right)\big|_{x=-0.5} \end{align}
Now, we can evaluate it using the geometric sum formula and then differentiate it, then evaluate it as $x=-0.5$.