Mathematical Induction (summation): $\sum^n_{k=1} k2^k =(n-1)(2^{n+1})+2$ I am stuck on this question from the IB Cambridge HL math text book about Mathematical induction. I am sorry about the bad formatting I am new and have no idea how to write the summation sign.
Using mathematical induction prove that the 
$$\sum^n_{k=1} k2^k =(n-1)(2^{n+1})+2$$ 
[correction made]
I tried solving it and got stuck on the let $n=k+1$ part 
So first I made $n=1$ and both sides equaled to $2$ 
then assume $n=k$ and got an expression which I don't know how to write here because of the formatting
then $n=K+1$ 
Thanks again 
 A: We need to prove that 
$$
\sum_{k=1}^nk2^k=(n-1)(2^{n+1})+2
$$
Consider $P_1$ where $k=1$. Left Hand Side (LHS) and Right Hand Side (RHS) are evaluated as follows. 
$$\sum_{k=1}^1k2^k=1\times2=2\quad\quad\quad\quad\quad\quad(1-1)(2^2)+2=2$$
Now, we assume that $P_m$ holds for some natural number $m$. (The trick here is that you will need to use this result later.)
$$
\sum_{k=1}^mk2^k=(m-1)(2^{m+1})+2
$$
It must be shown that $P_{m+1}$ holds too. Let us prove it. The RHS, which is relatively easier is
$$
\begin{align*}
[(m+1)-1](2^{m+2})+2&=m(2^{m+2})+2
\end{align*}
$$
What we need to do is to show that the LHS has this form:
$$
\begin{align*}
\sum_{k=1}^{m+1}k2^k&=\sum_{k=1}^mk2^k+(m+1)2^{m+1}\\
&=(m-1)(2^{m+1})+2+(m+1)2^{m+1}\\
&=(m-1)(2^{m+1})+(m+1)2^{m+1}+2\\&=(2m)(2^{m+1})+2\\
&=m(2^{m+2})+2
\end{align*}
$$
Since $P_1$ true, $P_m$ true $\rightarrow P_{m+1}$ true, by Mathematical Induction, $P_k$ true for $k=1, 2, \cdots$
Proven
A: Let $f(n) = (n-1)2^{n+1}+2$
and $g(n)=\sum^n_{k=1} k2^k$
Then
\begin{equation}\begin{aligned}
f(n+1) - f(n)&= (n) 2^{n+2}+2 - (n-1)2^{n+1}-2\\
&= (2n)2^{n+1} - (n-1)2^{n+1}\\
&= (n+1)2^{n+1}\\
&= g(n+1)-g(n)\\
\end{aligned}\end{equation}
Hence $g(n)=f(n) \Rightarrow g(n+1)=f(n+1)$
A: $\sum_{k=1}^{n}k2^{k}=2+(4+4)+(8+8+8)+...+(2^{n}+2^{n}+...+2^{n})$
$=(2+4+8+...+2^{n-1}+2^{n})+(4+8+...+2^{n-1}+2^{n})+(8+...+2^{n-1}+2^{n})+...+(2^{n-2}+2^{n-1}+2^{n})+(2^{n-1}+2^{n})+2^{n}$
$2(1+2+4+...+2^{n-2}+2^{n-1})+4(1+2+4+...+2^{n-3}+2^{n-2})+8(1+...+2^{n-3})+...+2^{n-2}(1+2+4)+2^{n-1}(1+2)+2^{n}$
$=2(2^{n}-1)+4(2^{n-1}-1)+8(2^{n-2}-1)+...+2^{n-2}(2^{3}-1)+2^{n-1}(2^{2}-1)+2^{n}$
$=(n-1)2^{n+1}+2^{n}-2-4-...-2^{n-1}=(n-1)2^{n+1}+2^{n}-2(2^{n-1}-1)$
$=(n-1)2^{n+1}+2$
This should be the formula. Now we prove this by induction. Both sides are $2$ at $n=1$. Assume it is true for $n\ge1$ and we show it for $n+1$.
$\sum_{k=1}^{n+1}k2^{k}=\sum_{k=1}^{n}k2^{k}+(n+1)2^{n+1}=(n-1)2^{n+1}+2+(n+1)2^{n+1}$
$=(2n)2^{n+1}+2=n2^{n+2}+2=((n+1)-1)2^{(n+1)+1}+2$.
A: Answer :
Hint:
Let it be true for k from 1 to n, where $$
\sum_{k=1}^nk2^k=(n-1)(2^{n+1})+2
$$
For k from 1 to n+1, 
$$
\sum_{k=1}^{n+1}k2^k=(n-1)(2^{n+1})+2 + (n+1)2^{n+1}$$
If you simplify, you will get
$$
\sum_{k=1}^{n+1}k2^k=(n+1-1)2^{(n+1+1)}+2
$$
Hence Proved
Thanks
Satish
A: Let :
$$S = \sum_{k = 1}^n k 2^k$$
We have :
$$\begin{array}{lcl}
S & = & \displaystyle \sum_{k = 0}^{n - 1} (k + 1) 2^{k + 1} \\[3mm]
& = & \displaystyle 2 \sum_{k = 1}^{n - 1} (k 2^k + 2^k) \\[3mm]
& = & \displaystyle 2 \sum_{k = 1}^{n - 1} k 2^k + 2 \sum_{k = 1}^{n - 1} 2^k \\[3mm]
& = & \displaystyle 2 \left(\sum_{k = 1}^n k 2^k - n 2^n\right) + \sum_{k = 0}^n 2^k \\[3mm]
& = & \displaystyle 2 \left(S - n 2^n\right) + \dfrac{2^{n + 1} - 2}{2 - 1} \\[3mm]
& = & \displaystyle 2 S - n 2^{n + 1} + 2^{n + 1} - 2 \\[3mm]
& = & \displaystyle 2 S - (n - 1) 2^{n + 1} - 2 \\[3mm]
\end{array}$$
We deduce that :
$$S = (n - 1) 2^{n + 1} + 2$$
