Show $\sum_{i=1}^{n+1} 2^ii = 2^{n+2}n+2$ , for all integers $n\ge 0$ using induction.

Question

How would i solve this problem?, i tried and went through the entire process is this correct and how would i factor the induction part.

$\sum_{i=1}^{n+1} i\cdot 2^i = n\cdot 2^{n+2}+2$ , for all integers $n\ge0.$

Would the expanded form be :

$1\cdot 2^1+2\cdot 2^2+3\cdot 2^2+\cdots +k\cdot 2^k+(k+1)\cdot 2^{k+1}$

Base case:

$1\cdot 2^1 =2,$

$0\cdot 2^{0+2}+2=2$

(k) term Assume that $k\cdot 2^k =k\cdot 2^{k+2}+2$ is true

(k+1) prove that $(k+1)\cdot 2^{k+1+2}+2$

induction

$(k\cdot 2^{k+2}+2)+(k+1)\cdot 2^{k+1}=(k+1)\cdot 2^{k+1+2}+2$

Answer 1

(k) term Assume that $k*2^k =k*2^{k+2}+2$ is true

No, you have to assume that $$\sum_{i=1}^{k+1} i\cdot 2^i = k\cdot 2^{k+2}+2$$

(k+1) prove that $(k+1)*2^{k+1+2}+2$

You have to prove that $$\sum_{i=1}^{(k+1)+1} i\cdot 2^i = (k+1)\cdot 2^{(k+1)+2}+2$$

induction $(k*2^{k+2}+2)+(k+1)*2^{k+1}=(k+1)*2^{k+1+2}+2$

This is wrong. See the red parts in the following : $$\begin{align}\sum_{i=1}^{(k+1)+1} i\cdot 2^i &= \left(\sum_{i=1}^{k+1} i\cdot 2^i \right)+(k+\color{red}{2})\cdot 2^{k+\color{red}{2}}\\&=(k\cdot 2^{k+2}+2)+(k+2)\cdot 2^{k+2}\\&=2^{k+2}(k+k+2)+2\\&=(k+1)\cdot 2^{k+3}+2\end{align}$$

Answer 2

Your last equality is actually false. Note that, in the $(k+1)$th case, the sum ends with $(k+2)2^{k+2}$. Thus, you actually want

$$(k2^{k+2}+2)+(k+2)2^{k+2}=(k+1)2^{k+1+2}+2$$

Start with the left hand side and factor out $2^{k+2}$. You get

$$\begin{align*} (k2^{k+2}+2)+(k+2)2^{k+2} & = 2^{k+2}(2+2k)+2 \\ & = (k+1)2^{k+1+2}+2 \end{align*}$$