+26367 Find the sum \(1*3 + 3*5 + 5*7 + 7*9 +\ldots\) for the first \(n\) terms.
\(\begin{array}{|rcll|} \hline \mathbf{1*3 + 3*5 + 5*7 + 7*9+ \ldots} &=& \mathbf{\sum \limits_{k=1}^{n} (2k-1)(2k+1)} \\ &=& \sum \limits_{k=1}^{n} (4k^2-1) \\ &=& \sum \limits_{k=1}^{n} 4k^2 - \sum \limits_{k=1}^{n} 1 \\ &=& 4\sum \limits_{k=1}^{n} k^2 - \sum \limits_{k=1}^{n} 1 \\ && \boxed{\sum \limits_{k=1}^{n} 1=n}\\ &=& 4\sum \limits_{k=1}^{n} k^2 - n \\ && \boxed{\sum \limits_{k=1}^{n} k^2=\dfrac{n(n+1)(2n+1)}{6} }\\\\ &=& \dfrac{4n(n+1)(2n+1)}{6} - n \\\\ &=& \dfrac{2n(n+1)(2n+1)}{3} - n \\\\ &=& \dfrac{2n(n+1)(2n+1)}{3} - \dfrac{3n}{3} \\\\ &=& \dfrac{n}{3}\Big( 2(n+1)(2n+1)-3 \Big) \\\\ &=& \dfrac{n}{3}\Big( (2n+2)(2n+1)-3 \Big) \\\\ &=& \dfrac{n}{3}\Big( 4n^2+6n+2-3 \Big) \\\\ \mathbf{1*3 + 3*5 + 5*7 + 7*9+ \ldots} &=& \mathbf{\dfrac{n}{3}\left( 4n^2+6n-1 \right)} \\ \hline \end{array}\)
+26367 Find the sum \(1*3 + 3*5 + 5*7 + 7*9 +\ldots\) for the first \(n\) terms.
\(\begin{array}{|rcll|} \hline \mathbf{1*3 + 3*5 + 5*7 + 7*9+ \ldots} &=& \mathbf{\sum \limits_{k=1}^{n} (2k-1)(2k+1)} \\ &=& \sum \limits_{k=1}^{n} (4k^2-1) \\ &=& \sum \limits_{k=1}^{n} 4k^2 - \sum \limits_{k=1}^{n} 1 \\ &=& 4\sum \limits_{k=1}^{n} k^2 - \sum \limits_{k=1}^{n} 1 \\ && \boxed{\sum \limits_{k=1}^{n} 1=n}\\ &=& 4\sum \limits_{k=1}^{n} k^2 - n \\ && \boxed{\sum \limits_{k=1}^{n} k^2=\dfrac{n(n+1)(2n+1)}{6} }\\\\ &=& \dfrac{4n(n+1)(2n+1)}{6} - n \\\\ &=& \dfrac{2n(n+1)(2n+1)}{3} - n \\\\ &=& \dfrac{2n(n+1)(2n+1)}{3} - \dfrac{3n}{3} \\\\ &=& \dfrac{n}{3}\Big( 2(n+1)(2n+1)-3 \Big) \\\\ &=& \dfrac{n}{3}\Big( (2n+2)(2n+1)-3 \Big) \\\\ &=& \dfrac{n}{3}\Big( 4n^2+6n+2-3 \Big) \\\\ \mathbf{1*3 + 3*5 + 5*7 + 7*9+ \ldots} &=& \mathbf{\dfrac{n}{3}\left( 4n^2+6n-1 \right)} \\ \hline \end{array}\)
