proofs by induction

**gohangoten1** · September 27th 2009, 11:03 AM

2+6+10+...+(4n-2)=2(n)^2

**galactus** · September 27th 2009, 11:19 AM

Show true for n=1: 4(1)-2=2(1)^2. True.

Assume true for $2+6+10+....+(4k-2)=2k^{2}$

Show true for $P_{k+1}$ :

$2+6+10+....+(4k-2)+(4k+2)=2k^{2}+4k+2=2(k+1)^{2}$

It is true for the $P_{k+1}$ case.

**gohangoten1** · September 27th 2009, 04:13 PM

Awsome

**gohangoten1** · September 27th 2009, 04:25 PM

2+4+6+....+2n = n(n+1)

**galactus** · September 27th 2009, 04:29 PM

Originally Posted by gohangoten1

2+4+6+....+2n = n(n+1)

Just add 2n+2 to both sides and see what that gives you on the right side.

I bet it's n(n+1)+2n+2=(n+1)(n+2).

Afterall, what would be the next number in the series?. 2n+2. See?

**gohangoten1** · September 27th 2009, 04:33 PM

5+10+15+...+5n= 5n(n+1)/2

**MacstersUndead** · September 27th 2009, 06:44 PM

Try this one on your own. All these induction proofs with sums follow the same pattern

Show for a base (usually n=1)
Suppose P(n) which is "5 + 10 + 15 + ... + 5n = 5n(n+1)/2"
Show P(n+1) supposing P(n)

with these sums, it amounts to this for P(n+1)

5 + 10 + 15 + ... + 5n + 5(n+1)
but what is the bolded part equal to according to assumption? can you finish?

**gohangoten1** · September 28th 2009, 05:29 AM

thanks

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