# Can someone guide me with this simple induction problem?

• Apr 24th 2010, 07:17 AM
matthayzon89
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(Headbang)

Thanks;)
• Apr 24th 2010, 08:19 AM
Quote:

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(Headbang)

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\$).
• Apr 24th 2010, 08:44 AM
baz
Quote:

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(Headbang)