Show that $\sum_{k=0}^n (k+1)x^k = \sum_{k=1}^{n+1} (k)x^{k+1}$. Note: for context, I have a 10-year old high-school math level and I'm trying to get back at math.
I have this exercise, asked in the form of:
The sum $\sum_{k=0}^n (k+1)x^k$ is equal to: (three possible solutions follow)
After spending about an hour on the problem, I looked up the solution and it only says that:
$$\sum_{k=0}^n (k+1)x^k = \sum_{k=1}^{n+1} (k)x^{k-1}$$
It is fairly evident to prove that both sides are equal by showing that they are both $$1 + 2x + 3x^2 + \cdots + (n+1)x^n$$
Now I'm wondering if I should have been able to find the right side of the equality just starting from the left side. Is this possible? Is it so evident that I'm just not finding it?
Thanks a lot!
Note: this exercise appears in H Prépa édition spéciale - Maths
 A: $$\sum_{k=0}^n (k+1)x^k = (0+1)+(1+1)x+(2+1)x^2+...+(n-1+1)x^{n-1}+(n+1)x^n$$
$$=1x^0+2x^1+3x^2+...+(n+1)x^{n}$$
$$\sum_{k=1}^{n+1} (k)x^\color{red}{k+1}=1x^2+2x^3+...+(n+1)x^{n+2}$$
so how are they the same?
Second formula should be 
$$\sum_{k=1}^{n+1} (k)x^\color{green}{k-1}=1x^{1-1}+2x^{2-1}+3x^{3-1}+...+(n+1)x^{n+1-1}$$
$$=1x^0+2x^1+3x^2+...+(n+1)x^{n}$$
A: Here is another way of looking at it.
Change of variable :    Put $k+1=u$
If $k=0$:              Then $u=1$
If $k=n$:              Then $u=n+1$
How does the equation look like now : 
$$\sum_{k=0}^n (k+1)x^k=\sum_{u=1}^{n+1} ux^{u-1} \text{ i.e., } \sum_{k=0}^n (k+1)x^k=\sum_{k=1}^{n+1} kx^{k-1}.$$
A: This can also be proved by using a change of indeces.
Let $$m=k+1$$
Then, when $k=0$, this implies that $m=k+1=0+1=1$.
Likewise, when $k=n$, this implies $m=n+1$.
Finally, since $m=k+1$, this implies that $k=m-1$
Hence,
$$\sum_{k=0}^n{\left(k+1\right)x^k}=\sum_{m=1}^{n+1}{\left(m-1+1\right)x^{m-1}}=\sum_{m=1}^{n+1}{\left(m\right)x^{m-1}}$$
Finally, since $m$ and $k$ are really arbitrary indices, you can replace $m$ with $k$ to get $$\sum_{k=1}^{n+1}{\left(k\right)x^{k-1}}$$
A: \begin{align}
\sum_{k=0}^5 (k+1)x^k & = 1x^0 + 2x^1+3x^2+4x^3+5x^4+6x^5 & & k \text{ goes from 0 to 5.} \\[12pt]
& = 1x^0 + 2x^1+3x^2+4x^3+5x^4+6x^5 = \sum_{k=1}^6 kx^{k-1} & & k\text{ goes from 1 to 6.}
\end{align}
A: Sum limits can be confusing when substituting, consider writting the limits as an inequality instead:
$$
\sum_{k=0}^n (k+1)x^k = \sum_{0 \leq k \leq n} (k+1)x^k = \sum_{0 \leq k-1 \leq n} ((k-1)+1)x^{k-1} = \sum_{1 \leq k \leq n+1} kx^{k-1} = \sum_{k=1}^{n+1} kx^{k-1}
$$
Only tricky part is now substituting the $k$ with $k-1$, but that is just a substitution $u=k+1$ followed by relabeling $u$ to $k$ again.
