Thread: Immediate help needed! Induction and Recursion!

Okay, here's a question, I don't think the right answer is there!

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

Which expression represents the given function?

(These are the options)

F(n)=n(n+1)

F(n)=n(n-1)

F(n)=2n(2n+1)

F(n)=(n-1)(n+1)

F(n)=n^2+2n

I thought at first that it was the last option, but I can't work it out just right..... I've now spent 2 hours on this question! and I need to go to bed soon! Thank you!

F(n)=2+4+6+...+2n is the same as F(n)=2(1+2+3+...+n)
The sum of the numbers from 1 to n for positive n = n*(n+1)/2.

So by substitution, F(n) = 2*n*(n+1) / 2 = n*(n+1), since the 2's cancel. This appears to be the first option in your list.

- Steve J

Originally Posted by Conorsmom

Okay, here's a question, I don't think the right answer is there!

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

Which expression represents the given function?

(These are the options)

F(n)=n(n+1)

F(n)=n(n-1)

F(n)=2n(2n+1)

F(n)=(n-1)(n+1)

F(n)=n^2+2n

I thought at first that it was the last option, but I can't work it out just right..... I've now spent 2 hours on this question! and I need to go to bed soon! Thank you!

To find the sum of terms in an arithmetic series, use the following formula.

In this formula:

Sn is the sum of the first n terms in a series
n is the number of terms you are adding up
a1 is the first term of the series
an is the nth term of the series
Here a1=2,n=n,an=2n,so the required sum F(n)is
F(n)=n(2+2n)/2=2n(n+1)/2=n(n+1) which is the first one in your list.