# Recurrence Relations

• Jul 28th 2011, 12:54 PM
lovesmath
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
• Jul 28th 2011, 01:07 PM
Plato
Re: Recurrence Relations
Quote:

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

$\displaystyle 3(1)+3(2)+\cdots+3(n)+n=3\frac{n(n+1)}{2}+n=~?$
• Jul 28th 2011, 02:16 PM
lovesmath
Re: Recurrence Relations
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?
• Jul 28th 2011, 02:25 PM
Plato
Re: Recurrence Relations
Quote:

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.
$\displaystyle 3+3(2)+2(3)+\cdots+3(n)+\color{red}n$.

That is not at all the same as $\displaystyle 3+2+4+6+8+...+2n$.

If you don't post correctly, how can we help? Do we guess?
• Jul 28th 2011, 02:40 PM
lovesmath
Re: Recurrence Relations
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.
• Jul 28th 2011, 02:50 PM
Plato
Re: Recurrence Relations
Quote:

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 $\displaystyle 3 + ... + 2n$ is so poorly written that no one can read it.

If you mean $\displaystyle 3 + 2+ 4+... + 2n$ then that is the same as
$\displaystyle 3+2(1)+2(2)+\cdots+2(n)=$
$\displaystyle 3+2(1+2+\cdots+n)=$
$\displaystyle 3+2\left(\frac{n(n+1)}{2}\right)$