Induction step completion

Question

I am confused how to go about the inductive step of the this given problem..

For all natural numbers $n, 2+4+6+....+2n =(n^2+n)$

I got the base case but now I am confused. Any help is greatly appreciated!

Proof: We proceed by induction.
          Base case: Set n = 1. Then 2 = (1)^2 + 1. Thus the 
          claim holds when n = 1.

       Inductive step: Let n>=1 be given.

Matthew Leingang · Answer 1 · 2017-11-15T18:32:32.593

0

The inductive step starts:

Let $n\geq 1$ be given such that $$ 2 + 4 + 6 + \dots + 2n = n^2 + n $$ We wish to show that $$ 2 + 4 + 6 + \dots + 2n + 2(n+1) = (n+1)^2 + (n+1) $$

Now you try to use the first equation to prove the second. Can you do that?

OP asks about the appearance of $2n$ in the second equation. If you infer the pattern from the $\cdots$, then $$ 2 + 4 + 6 + \dots + 2(n+1) = 2 + 4 + 6 + \dots + 2n + 2(n+1) $$ In both cases, you double all the numbers between $1$ and $n+1$, and add up those doubles. I just decided to explicitly write not only the last term in the series, but also the next-to-last term.

Why would I do that? It emphasizes that the left-hand side of the second equation is the same as the left-hand side of the first equation, with an additional term added on: $2(n+1)$. That is a hint for how to show the second equation is satisfied.

edited Nov 15 '17 at 18:32

answered Nov 15 '17 at 18:20

Matthew Leingang

26,112

Why did you keep the 2n and then add the 2(n+1)? Shouldn't the 2n have been just change to 2(n+1)? – Nov 15 '17 at 18:23
@MirrandaRyan Good question. See edit. – Matthew Leingang Nov 15 '17 at 18:32
@amWhy I think OP is confused about what is elided by the $\cdots$ in a series. In some sense both are valid expressions. – Matthew Leingang Nov 15 '17 at 18:34

S.S · Accepted Answer · 2017-11-15T18:49:06.083

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Add the (n+1)th term of the series on both sides of $P(n)$ : [Here i have assumed P(n) to be the the given proposition]$$2+4+6+...+2n+2(n+1)=n^2+n+2(n+1)$$

RHS: $n^2+n+2(n+1)=(n+1)^2+(n+1)$ [You may check that!] So as per induction hypothesis your proposition is true for $n+1$, ie $P(n+1)$ is true holding $P(n)$ to be true.That proves your proposition to be true for all natural numbers as per the standard induction hypothesis.

edited Nov 15 '17 at 18:49

answered Nov 15 '17 at 18:31

S.S

1,207

Okay so I understand why you added 2(n+1) to each side but how does that prove that it's true? – Nov 15 '17 at 18:41
Just assume the entire proposition in the question to be$P(n)$ and then just plug in n+1 there, you get $P(n+1)$ which is actually what i have deduced above. We actually had assumed $P(n)$ to be true and then holding that we proved $P(n+1)$ to be true. That is what we wanted. – S.S Nov 15 '17 at 18:46

Induction step completion

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