Calculus Calculus Math Forum

February 27th, 2015, 05:31 AM   #1
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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 10:48 AM.

 February 27th, 2015, 10:18 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 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?
February 28th, 2015, 10:10 AM   #3
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Quote:
 Originally Posted by MarkFL 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!

 February 28th, 2015, 11:17 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 895 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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