February 27th, 2015, 04:31 AM  #1 
Senior Member Joined: Jan 2015 From: USA Posts: 107 Thanks: 2  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! Last edited by skipjack; February 27th, 2015 at 09:48 AM. 
February 27th, 2015, 09:18 AM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 511 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, 09:10 AM  #3  
Senior Member Joined: Jan 2015 From: USA Posts: 107 Thanks: 2  Quote:
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, 10:17 AM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
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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addition, cuadratic, quadratic, series 
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