Is there any. elementary formula for the sequence$\sum_{k=1}^{n}\left(2k-1\right)\left(\frac{1}{2}\right)^{k}$ Is there any formula for the following sequence which does not use any derivative and also is less 
advanced:
$$\sum_{k=1}^{n}\left(2k-1\right)\left(\frac{1}{2}\right)^{k}$$
I've calculated the general formula 
and here is a photo of the solution
 1: 
 A: We have that
$$\sum_{k=1}^{n}\left(2k-1\right)\left(\frac{1}{2}\right)^{k}=2\sum_{k=1}^{n}k\left(\frac{1}{2}\right)^{k}-\sum_{k=1}^{n}\left(\frac{1}{2}\right)^{k}$$
then for the first term we can use the approch indicated here


*

*Sum of a power series $n x^n$
A: It's a fairly simple proof by induction.
$$\mathcal{H}_n \ : \ \left\{ \sum_{k=1}^{n}\left(2k-1\right)\left(\frac{1}{2}\right)^{k} = -2^{-n}(2n+3)+3\right\}$$
Base case is obvious, $\mathcal{H}_1$ is true.
Induction. Suppose $\mathcal{H}_n$ true. Now
$$\sum_{k=1}^{n+1}\left(2k-1\right)\left(\frac{1}{2}\right)^{k} = \sum_{k=1}^{n}\left(2k-1\right)\left(\frac{1}{2}\right)^{k} +(2n+1)\left(\frac{1}{2}\right)^{n+1}$$
By induction
$$\begin{align*}
\sum_{k=1}^{n+1}\left(2k-1\right)\left(\frac{1}{2}\right)^{k} &= -2^{-n}(2n+3) +3+(2n+1)\left(\frac{1}{2}\right)^{n+1}\\
&= \underbrace{-2^{-n}(2n+1)}_{A} -\underbrace{2^{-n}2}_{B} +\underbrace{(2n+1)2^{-n}\left(\frac{1}{2}\right)}_{C} + 3\\
\end{align*}
$$
Summing $A$ and $C$, and rewriting $B$ as $2^{-n}2=2^{-(n+1)}4$ we get
$$\sum_{k=1}^{n+1}\left(2k-1\right)\left(\frac{1}{2}\right)^{k} = -2^{-(n+1)}(2n+5)+3$$
Hence $\mathcal{H}_{n+1}$ is true, and by induction principle,
$$\forall n\geq 1, \  \sum_{k=1}^{n}\left(2k-1\right)\left(\frac{1}{2}\right)^{k} = -2^{-n}(2n+3)+3$$
Edit - more direct approach Using inspiration from this question
Let 
$$F_n(x)=\sum_{k=1}^nkx^k=x+2x^2+\ldots+nx^n$$
Then
$$F_n(x)-xF_n(x) = x+x^2+x^3+\ldots+x^n-nx^{n+1}$$
$$F_n(x)-xF_n(x) = \sum_{k=1}^nx^k-nx^{n+1}$$
$$F_n(x)-xF_n(x) = x\frac{1-x^n}{1-x}-nx^{n+1}$$
So that
$$F_n(x) = x\frac{1-x^n}{(1-x)^2}-\frac{nx^{n+1}}{1-x}$$
Now, using @user remark, we have
$$\sum_{k=1}^{n}\left(2k-1\right)\left(\frac{1}{2}\right)^{k}=2\sum_{k=1}^{n}k\left(\frac{1}{2}\right)^{k}-\sum_{k=1}^{n}\left(\frac{1}{2}\right)^{k}$$
And
$$\sum_{k=1}^{n}(2k-1)\left(\frac{1}{2}\right)^{k}= 2F_n(1/2)-(1-2^{-n})$$
Using our formula for $F_n(x)$,
$$\sum_{k=1}^{n}(2k-1)\left(\frac{1}{2}\right)^{k}= 2\cdot\left(\frac{1}{2}\frac{1-2^{-n}}{(1/2)^2}-\frac{n2^{-(n+1)}}{1/2}\right) - (1-2^{-n})$$
$$\sum_{k=1}^{n}(2k-1)\left(\frac{1}{2}\right)^{k}= 4 (1-2^{-n}) - 4n2^{-(n+1)} - (1-2^{-n})$$
$$\sum_{k=1}^{n}(2k-1)\left(\frac{1}{2}\right)^{k}= 3 (1-2^{-n}) - 4n2^{-(n+1)}$$
And finally
$$\sum_{k=1}^{n}\left(2k-1\right)\left(\frac{1}{2}\right)^{k} = -2^{-n}(2n+3)+3$$
