value of $\sum_{k=0}^{49}(k+1)(1.06)^{k+1}$ How do I calculate the value of $\sum_{k=0}^{49}(k+1)(1.06)^{k+1}$? I do not know the way to solve this type of a summation. Any guidance will be much appreciated 
 A: For $x\in\mathbb{R}\setminus\{1\}$,
$$\begin{align*}
\sum_{k=0}^{49}(k+1)x^{k+1}&=x\sum_{k=0}^{49}(k+1)x^{k}=x\frac{d}{dx}\sum_{k=0}^{49}x^{k+1} \\
&=x\frac{d}{dx}\sum_{k=1}^{50}x^{k} =x\frac{d}{dx}\left(x\frac{x^{50}-1}{x-1}\right)
\end{align*}$$
Now, you can easily compute the last expression, and evaluate it at $x=1.06$.
A: Hint: If you know the value of $\sum\limits_{k=0}^nkx^k$ for $n=50$ and $x=1.06$, you are done. This is almost the derivative of $S_n(x)=\sum\limits_{k=0}^nx^k$, right? So, if you know $S_n(x)$ and how to differentiate it, you are done. But $S_n(x)$ is simply a geometric series hence...
A: $$\sum_{k=1}^n kx^k=x\frac{d}{dx}\sum_{k=1}^n x^k=x\frac{d}{dx}\left(\frac{x-x^{n+1}}{1-x}\right)=\frac{x(nx^{n+1}+1-(n+1)x^n)}{(1-x)^2}$$
$$ \sum_{k=0}^{49}(k+1)(1.06)^{k+1}= \sum_{k=1}^{50}k(1.06)^{k}$$
A: Let $f(x) = \sum_{k=1}^N k x^{k-1}$. Consider $$F(x) = \int f(x) = \sum_{k=1}^N \int kx^{k-1}dx = \sum_{k=1}^N x^k = \frac{1-x^N}{1-x} $$
Thus we have $$f(x) = F'(x) = \frac{-Nx^{N-1}(1-x) + (1-x^N)}{(1-x)^2}$$ 
Plug that into your series and you can get the answer. 
A: $$\sum_{k=0}^nx^k=\frac{1-x^{n+1}}{1-x}$$
$$\sum_{k=1}^nkx^{k-1}=\sum_{k=0}^{n-1}(k+1)x^{k}=\left(\frac{1-x^{n+1}}{1-x}\right)'$$
$$\sum_{k=0}^{n-1}(k+1)x^{k+1}=x\left(\frac{1-x^{n+1}}{1-x}\right)'$$
