Using induction to prove $\sum_{i = 1}^{n} i\cdot 2^i = (n - 1) \cdot 2^{n + 1} + 2$

Question

I have to prove the following proposition:

$$\sum_{i = 1}^{n} i\cdot 2^i = (n - 1) \cdot 2^{n + 1} + 2$$

I have the following proof thus far:

Proof by Induction

Basis Step $(n = 1)$

LHS: $\sum_{i = 1}^{n} i2^i = 1 \cdot 2^1 = 2$

RHS: $2^2(0) + 2 = 2$

LHS = RHS

Inductive Hypothesis

Assume that the proposition is true for $i = 1, 2, 3, \ldots, k$

Inductive Step $(n = k + 1)$

LHS: $\sum_{i = 1}^{k} i2^i + (k + 1) \cdot 2^{k + 1} = 1 \cdot 2^1 + 2 \cdot 2^2 + \ldots + k \cdot 2^k + (k + 1) \cdot 2^{k + 1}$

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

LHS = RHS

QED

Am I correct? If not, am I headed in the right direction? I understand the basics of induction, but am continuing to apply it. Thank you!

Answer 1

Your base case is correct, but you need more information in the inductive hypothesis.

Base Case: $(n = 1)$ $$\sum_{i=1}^{1}i2^i = 1\cdot 2^1 = 2 = (1-1) \cdot 2^{1+1} + 2 $$ Inductive Hypothesis: Assume that $$\sum_{i=1}^{k} i2^i = (k-1) \cdot 2^{k+1} + 2 \hspace{1in} \text{IH}$$ for some integer $k >1$. You must show that $$\sum_{i=1}^{k+1} i2^i = k2^{k+2} + 2$$ Inductive Step: $$\sum_{i=1}^{k+1} i2^i = \sum_{i=1}^{k} i2^i+ \sum_{i=k+1}^{k+1} i2^i = (k-1) \cdot 2^{k+1} + 2 + (k+1)\cdot 2^{k+1} \hspace{.5in} \text{By IH}$$ $$= (k-1)\cdot 2^{k+1} + 2$$ $$= 2^{k+1} [ (k-1) + (k+1)] + 2$$ $$= 2^{k+1}\cdot 2k + 2$$ $$= k2^{k+1}\cdot 2^1 + 2$$ $$= k2^{k+2} + 2$$ Thus, $\sum_{i=1}^{n} i2^i = (n-1) \cdot 2^{n+1} + 2 \ \text{for any } n\geq 1$

QED