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Find the sum 1*3 + 3*5 + 5*7 + ... for the first n terms.

 Jan 14, 2021

Best Answer 

 #1
avatar+25657 
+3

Find the sum \(1*3 + 3*5 + 5*7 + 7*9 +\ldots\) for the first \(n\) terms.

 

\(\begin{array}{|rcll|} \hline \mathbf{1*3 + 3*5 + 5*7 + 7*9+ \ldots} &=& \mathbf{\sum \limits_{k=1}^{n} (2k-1)(2k+1)} \\ &=& \sum \limits_{k=1}^{n} (4k^2-1) \\ &=& \sum \limits_{k=1}^{n} 4k^2 - \sum \limits_{k=1}^{n} 1 \\ &=& 4\sum \limits_{k=1}^{n} k^2 - \sum \limits_{k=1}^{n} 1 \\ && \boxed{\sum \limits_{k=1}^{n} 1=n}\\ &=& 4\sum \limits_{k=1}^{n} k^2 - n \\ && \boxed{\sum \limits_{k=1}^{n} k^2=\dfrac{n(n+1)(2n+1)}{6} }\\\\ &=& \dfrac{4n(n+1)(2n+1)}{6} - n \\\\ &=& \dfrac{2n(n+1)(2n+1)}{3} - n \\\\ &=& \dfrac{2n(n+1)(2n+1)}{3} - \dfrac{3n}{3} \\\\ &=& \dfrac{n}{3}\Big( 2(n+1)(2n+1)-3 \Big) \\\\ &=& \dfrac{n}{3}\Big( (2n+2)(2n+1)-3 \Big) \\\\ &=& \dfrac{n}{3}\Big( 4n^2+6n+2-3 \Big) \\\\ \mathbf{1*3 + 3*5 + 5*7 + 7*9+ \ldots} &=& \mathbf{\dfrac{n}{3}\left( 4n^2+6n-1 \right)} \\ \hline \end{array}\)

 

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 Jan 14, 2021
 #1
avatar+25657 
+3
Best Answer

Find the sum \(1*3 + 3*5 + 5*7 + 7*9 +\ldots\) for the first \(n\) terms.

 

\(\begin{array}{|rcll|} \hline \mathbf{1*3 + 3*5 + 5*7 + 7*9+ \ldots} &=& \mathbf{\sum \limits_{k=1}^{n} (2k-1)(2k+1)} \\ &=& \sum \limits_{k=1}^{n} (4k^2-1) \\ &=& \sum \limits_{k=1}^{n} 4k^2 - \sum \limits_{k=1}^{n} 1 \\ &=& 4\sum \limits_{k=1}^{n} k^2 - \sum \limits_{k=1}^{n} 1 \\ && \boxed{\sum \limits_{k=1}^{n} 1=n}\\ &=& 4\sum \limits_{k=1}^{n} k^2 - n \\ && \boxed{\sum \limits_{k=1}^{n} k^2=\dfrac{n(n+1)(2n+1)}{6} }\\\\ &=& \dfrac{4n(n+1)(2n+1)}{6} - n \\\\ &=& \dfrac{2n(n+1)(2n+1)}{3} - n \\\\ &=& \dfrac{2n(n+1)(2n+1)}{3} - \dfrac{3n}{3} \\\\ &=& \dfrac{n}{3}\Big( 2(n+1)(2n+1)-3 \Big) \\\\ &=& \dfrac{n}{3}\Big( (2n+2)(2n+1)-3 \Big) \\\\ &=& \dfrac{n}{3}\Big( 4n^2+6n+2-3 \Big) \\\\ \mathbf{1*3 + 3*5 + 5*7 + 7*9+ \ldots} &=& \mathbf{\dfrac{n}{3}\left( 4n^2+6n-1 \right)} \\ \hline \end{array}\)

 

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heureka Jan 14, 2021
 #2
avatar+117447 
+1

Nice work, heureka  !!!!

 

cool cool cool

CPhill  Jan 14, 2021
 #3
avatar+25657 
+1

Thank you CPhill !

 

laugh

heureka  Jan 14, 2021
edited by heureka  Jan 14, 2021

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