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https://web2.0calc.com/questions/help-me-please_3921

 

Compute the sum \((a +(2n+1)d)^2- (a + (2n)d)^2 +(a + (2n-1)d)^2 - (a+(2n-2)d)^2 + \cdots + (a+d)^2 - a^2.\)

 

Please write all of the steps! There was another post on this question, but I didn't get how the guest got the answer.

 Jun 25, 2020
 #1
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See Alan's answer here:    https://web2.0calc.com/members/alan/?answerpage=751

Note: If you don't understand Alan's answer, there isn't anybody here, as far as I can tell, that understands it better than Alan. If you are looking for a "simpler" solution, I'm afraid you may have to find it somewhere else.

I suggest that you interact with your student friends and with your teacher to arrive at some sort of a solution that you will understand. Even WolframAlpha comes up with this message: "(unable to determine general term)" ! Good luck to you.

 Jun 25, 2020
edited by Guest  Jun 25, 2020
 #2
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Group the terms in pairs. Using the formula a2 - b2 = (a - b)(a + b), we have

 

\(\quad(a +(2n+1)d)^2- (a + (2n)d)^2 +(a + (2n-1)d)^2 - (a+(2n-2)d)^2 + \cdots + (a+d)^2 - a^2\\ =\left((a +(2n+1)d)^2- (a + (2n)d)^2\right) +\left((a + (2n-1)d)^2 - (a+(2n-2)d)^2\right) + \cdots + \left((a+d)^2 - a^2\right)\\ =(\color{red}(a + (2n + 1)d)\color{black} - \color{blue}(a + (2n)d)\color{black})(\color{red}(a + (2n + 1)d)\color{black} + \color{blue}(a + (2n)d)\color{black}) \\\quad\quad+ (\color{red}(a + (2n-1)d)\color{black} - \color{blue}(a + (2n-2)d)\color{black})(\color{red}(a + (2n-1)d)\color{black} + \color{blue}(a + (2n-2)d)\color{black}) + \cdots \\\quad\quad\qquad+(\color{red}(a + d)\color{black} - \color{blue}a\color{black})(\color{red}(a + d)\color{black} + \color{blue}a\color{black})\)

 

The last line may be a bit long, but if we simplify it:

 

\(= d(2a + \color{red}(4n + 1)\color{black}d) + d(2a + \color{red}(4n - 3)\color{black}d) + \cdots + d(2a + \color{red}1\color{black}d)\)

 

Now we see that the original expression is nothing other than the sum of A.S. in disguise.

 

Factorising out a common factor for a more easy-to-spot A.S.:

 

\(= d((2a + (4n + 1)d) + (2a + (4n - 3)d) + \cdots + (2a + d))\)

 

We can now see the expression inside the bracket is an A.S. with first term = (2a + (4n + 1)d), common difference = -4d, number of terms = (n + 1).

 

We can plug those into the formula:

 

\(\quad\text{sum of A.S.} \\= \dfrac{\text{number of terms}}{2} \left(2\times \text{first term} + (\text{number of terms} - 1)\times \text{common difference}\right) \\= \dfrac{n + 1}{2}\left(2(2a + (4n + 1)d) + n(-4d)\right) \\= \dfrac{n + 1}2\left(4a + (4n + 2)d\right) \\= (n + 1)(2a+(2n + 1)d)\)

 Jun 26, 2020
 #3
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Thank you very much Max. You are as brilliant as ever. Congrats.

 Jun 26, 2020

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