Thread: Recurrence Relations

The formula 1+2+3+...+n = (n(n+1))/2 is true for all integers n>=1. Use this fact to solve each of the following problems:
If n is an integer and n>=1, find a formula for the expression
3+3(2)+3(3)+...+3(n)+n

Is this correct?
3+3((n(n+1))/2)+3
3+(3/2)(n(n+1))+3
3+(3/2)(n^2+n)+3

Originally Posted by lovesmath

The formula 1+2+3+...+n = (n(n+1))/2 is true for all integers n>=1. Use this fact to solve each of the following problems: If n is an integer and n>=1, find a formula for the expression
3+3(2)+3(3)+...+3(n)+n

The answer to 3+2+4+6+8+...+2n (if n is an integer and n>=1) is n^2+n+3. I thought since the original question I posted began with a 3, it would also add 3 like this example. If not, why?

Originally Posted by lovesmath

The answer to 3+2+4+6+8+...+2n (if n is an integer and n>=1) is n^2+n+3. I thought since the original question I posted began with a 3, it would also add 3 like this example. If not, why?

Look at what you did post.
.

That is not at all the same as .

If you don't post correctly, how can we help? Do we guess?

I was using the 3 + ... + 2n as an example. Why is it 3 + 2((n(n+1))2) = n^2 + n + 3? I don't understand where the 3 came from other than the fact that it is the first term in the sequence. I used that as an example because my problem, 3 + ... + 3n + n, also starts with a 3.

Originally Posted by lovesmath

I was using the 3 + ... + 2n as an example. Why is it 3 + 2((n(n+1))2) = n^2 + n + 3? I don't understand where the 3 came from other than the fact that it is the first term in the sequence. I used that as an example because my problem, 3 + ... + 3n + n, also starts with a 3.

The expression is so poorly written that no one can read it.

If you mean then that is the same as