# Thread: proofs by induction

1. ## proofs by induction

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

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

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

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

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

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

Awsome

4. ## got another

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

5. 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?

6. ## got another

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

7. 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?

thanks