Mathematical Induction Problem, Problem with Algebra in the Proof

Question

I have the statement: $$2+6+16+\cdots +(n+1)\cdot2^{n-1} = n\cdot 2^n$$ Assume the above statement is true, $P(n)$. Now I'll show $P(n+1)$ is true. $$2 +6+16+\cdots+(n+1)\cdot2^{n-1} + ((n+1)+1)\cdot2^n = (n+1)\cdot2^{n+1} \\ \text{Prove the left side to equal the right side.} \\ (n+1)\cdot2^{n-1} + (n+2) \cdot2^n \\ n\cdot2^{n-1} + 2^1\cdot2^{n-1} +(n+2)\cdot2^n \\ n+2^{n-1} + 2^n +(k+2)\cdot2^n$$ I am stuck on the last part. I don't know how to proceed with the algebra. What should I do to make this left side of the statement equal the right side? If I made nay pervious mistake in the algebra, my mistake. Did all of calculus and I still struggle with some simple algebra.

Answer 1

If $\sum_{k=1}^n a_k = f(n)$ then $\sum_{k=1}^{n+1} a_k = (\sum_{k=1}^n a_k ) + a_{n+1} = f(n) + a_{n+1}$.

So if

$2+6+16+\cdots +(n+1)\cdot2^{n-1} = n\cdot 2^n$ then

$2+6+16+\cdots +(n+1)\cdot2^{n-1} + (n+2)\cdot 2^n = n\cdot 2^n + (n+2)\cdot 2^n$

So we have to prove $n\cdot 2^n + (n+2)\cdot 2^n = (n+1)2^{n+1}$. Can we?

That should be straightforward.

$n\cdot 2^n + (n+2)\cdot 2^n = [n + (n+2)]2^n= =(2n+2)2^n =2(n+1)2^n=(n+1)2^{n+1}$.

That's it.

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Note: To prove $a_1 + a_2 + ..... + a_n = f(n)$ by induction, it is always sufficient to prove:

1. $a_1 = f(1)$

2. $f(n) + a_{n+1} = f(n+1)$.

Always. If you can do those, that proof by induction will always work. That should always be your first strategy.

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Correction

$$\color{blue}{2 +6+16+\cdots+(n+1)\cdot2^{n-1}} + ((n+1)+1)\cdot2^n = (n+1)\cdot2^{n+1} \\ \text{Prove the left side to equal the right side.} \\ \color{blue}{n\cdot2^n} + (n+2) \cdot2^n \\ (2n+2)2^n \\ (n+1)\cdot2^{n+1}$$

Answer 2

You assume that $2+6+16+\cdots+(n+1)\cdot2^{n-1}=n\cdot2^n$,

so you want to show that $n\cdot2^n+(n+2)2^n=(n+1)2^{n+1}$.

Well, that holds because $n\cdot2^n\cdot2+2\cdot2^n=n\cdot2^{n+1}+2^{n+1}$.