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February 27th, 2015, 04:31 AM   #1
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Series addition of a quadratic

Hi! I don't know how to solve this problem. Any ideas?

In case you can't see well the numbers in the image, it's the sum from k=1 to 11 of (k+1) (k+2).
Thanx!
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Last edited by skipjack; February 27th, 2015 at 09:48 AM.
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February 27th, 2015, 09:18 AM   #2
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Math Focus: Calculus/ODEs
We are given the sum:

$\displaystyle S=\sum_{k=1}^n\left((k+1)(k+2)\right)$

Expanding the summand, we obtain:

$\displaystyle S=\sum_{k=1}^n\left(k^2+3k+2\right)$

Using the linearity of sums, we may write:

$\displaystyle S=\sum_{k=1}^n\left(k^2\right)+3\sum_{k=1}^n\left( k\right)+2\sum_{k=1}^n\left(1\right)$

Now, there are 3 formulas we may use:

$\displaystyle \sum_{k=1}^n\left(1\right)=n\tag{1}$

$\displaystyle \sum_{k=1}^n\left(k\right)=\frac{n(n+1)}{2}\tag{2}$

$\displaystyle \sum_{k=1}^n\left(k^2\right)=\frac{n(n+1)(2n+1)}{6 }\tag{3}$

Can you proceed?
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February 28th, 2015, 09:10 AM   #3
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Quote:
Originally Posted by MarkFL View Post
We are given the sum:

$\displaystyle S=\sum_{k=1}^n\left((k+1)(k+2)\right)$

Expanding the summand, we obtain:

$\displaystyle S=\sum_{k=1}^n\left(k^2+3k+2\right)$

Using the linearity of sums, we may write:

$\displaystyle S=\sum_{k=1}^n\left(k^2\right)+3\sum_{k=1}^n\left( k\right)+2\sum_{k=1}^n\left(1\right)$

Now, there are 3 formulas we may use:

$\displaystyle \sum_{k=1}^n\left(1\right)=n\tag{1}$

$\displaystyle \sum_{k=1}^n\left(k\right)=\frac{n(n+1)}{2}\tag{2}$

$\displaystyle \sum_{k=1}^n\left(k^2\right)=\frac{n(n+1)(2n+1)}{6 }\tag{3}$

Can you proceed?

I understand! However, I dont know how you managed to generate "An" for the 3 series. Is there a formula or was it by trial and error? I know there are formulas for arithmetic and geometric series, but these are neither arithmetic nor geometric. Thanx!
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February 28th, 2015, 10:17 AM   #4
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The original problem involved (k+ 1)(k+ 2). MarkFL multiplied to get $\displaystyle k^2+ 3k+ 2$. No, he didn't use "trial and error" to break that into three series, he just separated "powers of k": $\displaystyle k^2$, $\displaystyle 3k$, and $\displaystyle 2$.
And he did that because he (and, hopefully you) knows formulas for such things as $\displaystyle \sum 1$, $\displaystyle \sum k$, and $\displaystyle \sum k^2$.
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