# Thread: Can someone guide me with this simple induction problem?

1. ## Can someone guide me with this simple induction problem?

Prove: 2 + 4 + 6 + ... + 2n= n^2+n for all integers n >= 1.

Proof:

P(n):2+4+6+...+2n= n^2+n

P(2):2+4+6+...+2n= 2^2+2= 6 which is true.

Suppose that P(k): 2+4+6+...+2k= k^2+k where k>=1.

2+4+6+...+2k+(2k+1)= (2k+1)^2+(2k+1)= 4k^2+2k+2k+1 +(2k + 1)= 4k^2+6k+2= 2(2k^2+3k+2)= This is as far as I have gotten

Thanks

2. Originally Posted by matthayzon89
Prove: 2 + 4 + 6 + ... + 2n= n^2+n for all integers n >= 1.

Proof:

P(n):2+4+6+...+2n= n^2+n

P(2):2+4+6+...+2n= 2^2+2= 6 which is true.

Suppose that P(k): 2+4+6+...+2k= k^2+k where k>=1.

2+4+6+...+2k+(2k+1)= (2k+1)^2+(2k+1)= 4k^2+2k+2k+1 +(2k + 1)= 4k^2+6k+2= 2(2k^2+3k+2)= This is as far as I have gotten

Thanks
Assume true for $\displaystyle P(k)$

Then $\displaystyle P(k+1) = P(k) + 2(k+1) = k^2 + k + 2k + 2$.

$\displaystyle = k^2 + 2k + 1 + (k+1) = (k+1)^2 + (k+1)$.

In general with these induction things usually the $\displaystyle k+1$ step involves saying $\displaystyle P(k+1) = P(k) +$ ... Then sub in what P(k) equals (not the sequence though, in this case $\displaystyle P(k) = 2^k + k$).

3. Originally Posted by matthayzon89
Prove: 2 + 4 + 6 + ... + 2n= n^2+n for all integers n >= 1.

Proof:

P(n):2+4+6+...+2n= n^2+n

P(2):2+4+6+...+2n= 2^2+2= 6 which is true.

Suppose that P(k): 2+4+6+...+2k= k^2+k where k>=1.

2+4+6+...+2k+(2k+1)= (2k+1)^2+(2k+1)= 4k^2+2k+2k+1 +(2k + 1)= 4k^2+6k+2= 2(2k^2+3k+2)= This is as far as I have gotten