# Function is F(n) = n(n+1)

1. Oct 15, 2008

### goofyfootsp

F(n) = 2+4+6=...+2n
I know the expression that represents the given function is F(n) = n(n+1),

my question is F(n) = 3+5+7...2n the same F(n) = n(n+1)< if not can anyone expain?

Thanks

2. Oct 15, 2008

### mathman

Re: Recursion

Please correct! You have a sequence of odd numbers, but the end term is even!

3. Oct 16, 2008

### robert Ihnot

Re: Recursion

This was gone into from General Math: Simple Sequences. The nth term is

$$\sum_1^n (2j-1)=n^2.$$ Or: 1=1, 1+3=2^2, 1+3+5 = 3^2, etc...

Last edited: Oct 16, 2008
4. Oct 16, 2008

### HallsofIvy

Re: Recursion

Why should they have the same sum when you are summing different numbers?

It is not to difficult to show that if you have any arithmetic sequence: an= a+ id, where a and d are fixed and i ranges from 1 to n, sums to n times the average of a1 and an: n*(a+ d+ (a+ nd))/2= n(a+ (d/2)(n+1)).

In the case of 2+ 4+ 6+ ...+ 2n, a= 0 and d= 2. The sum is n(0+ (2/2)(n+1)= n(n+1)
In the case of 1+ 3+ 5+ ...+ 2n+1, a= -1 and d= 2. The sum is n(-1+ (2/2)(n+1))= n2.

5. Oct 16, 2008

### goofyfootsp

Re: Recursion

Thank you all for your help in explaining this, I appreciate it greatly.