Prove using mathematical induction $1\cdot2+2\cdot2^2+3\cdot2^3+\ldots+n\cdot2^n=2[1+(n-1)2^n]$

Question

Prove the result using Mathematical Induction $$1\cdot2+2\cdot2^2+3\cdot2^3+\ldots\ldots+n\cdot2^n=2[1+(n-1)2^n].$$

I've been stuck on this problem for hours, I have no idea how do even calculate it. The exponents throw me off. If anyone can help me break it down step-by-step, I would truly appreciate it.

Here's my attempt

Answer 1

From your points $2$ and $3$, we need to show that

$$2(1+(k-1)2^k)+(k+1)2^{k+1}= 2(1+k2^{k+1})\\\stackrel{\text{divide by 2}}\iff 1+(k-1)2^k+(k+1)2^{k}= 1+k2^{k+1}\\\stackrel{\text{cancel out 1}}\iff (k-1)2^k+(k+1)2^{k}= k2^{k+1}\\\stackrel{\text{divide by} \,2^k}\iff (k-1)+(k+1)= 2k$$

Answer 2

Without induction it is rather simple problem.

$$\sum_{k=1}^n kx^k=x\sum_{k=1}^n kx^{k-1}=x\sum_{k=1}^n(x^k)'=x\left(\sum_{k=1}^n x^k\right)'=x\left(x\frac{x^n-1}{x-1}\right)'.$$ Now compute the derivative and put $x=2.$

Answer 3

Hint:

The base case is fairly easy: $0\times\left(\ldots\right)=2\left(1+(0-1)\times2^0\right)$. If we assume that $\sum_{n=0}^k n2^{n}=2\left(1+(k-1)\cdot2^{k}\right)$, then our goal for the induction case is to prove that: $$ \underbrace{2\left(1+(k-1)\cdot2^{k}\right)}_{\sum_{n=0}^k {n2^{n}}}+ \underbrace{(k+1)2^{k+1}}_{\text{$(k+1)$'th term}}=2\left(1+k\cdot2^{k+1}\right)$$

Then, divide through by 2 and expand. Can you take it from here?

Answer 4

You've made a mistake while doubling $2[1+(K-1)2^K]$.

To give a clearer and more concise presentation of the inductive step, I suggest writing one side at one row.

\begin{align} & 2[1+(K-1)2^K] + (K+1) 2^{K+1} \\ &= 2 + (K-1) 2^{K+1} + (K+1) 2^{K+1} \\ &= 2 + 2K 2^{K+1} \\ &= 2[1 + K 2^{K+1}] \end{align}