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Find (1^2 + 1*2 + 2^2) + (2^2 + 2*3 + 3^2) + ... + (99^2 + 99*100 + 100^2).

 Dec 4, 2019
 #1
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Find \((1^2 + 1*2 + 2^2) + (2^2 + 2*3 + 3^2) + \ldots + (99^2 + 99*100 + 100^2)\).

 

\(\begin{array}{|rcll|} \hline && \mathbf{(1^2 + 1*2 + 2^2) + (2^2 + 2*3 + 3^2) + \ldots + (99^2 + 99*100 + 100^2)} \\ &=& \sum \limits_{n=1}^{99} \left( n^2+n(n+1)+(n+1)^2 \right) \\ &=& \sum \limits_{n=1}^{99} \left( n^2+n^2+n +n^2+2n+1 \right) \\ &=& \sum \limits_{n=1}^{99} \left( 3n^2+3n+1 \right) \\ &=& \sum \limits_{n=1}^{99} \left( 3n^2 \right) +\sum \limits_{n=1}^{99} \left( 3n\right) +\sum \limits_{n=1}^{99} \left( 1 \right) \\ &=& 3\sum \limits_{n=1}^{99} \left( n^2 \right) +3\sum \limits_{n=1}^{99} \left( n\right) + \underbrace{\sum \limits_{n=1}^{99} \left( 1 \right)}_{= 99*1=99} \\ &=& 3\sum \limits_{n=1}^{99} \left( n^2 \right) +3\sum \limits_{n=1}^{99} \left( n\right) + 99 \\\\ && \boxed{\sum \limits_{n=1}^m (n^2) = \dfrac{m(m+1)(2m+1)}{6} \\ \sum \limits_{n=1}^m (n ) = \dfrac{m(m+1)}{2} } \\\\ &=& 3 *\left( \dfrac{99*(99+1)*(2*99+1)}{6}\right) +3 *\left( \dfrac{99*(99+1)}{2}\right) + 99 \\ &=& 3 *\left( \dfrac{99*100*199}{6}\right) +3 *\left( \dfrac{99*100}{2}\right) + 99 \\ &=& 99*50*199 + 3 * 99* 50 + 99 \\ &=& \mathbf{999999} \\ \hline \end{array}\)

 

laugh

 Dec 4, 2019
 #2
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∑[ (3 n^2 + 3 n + 1), n, 1, 99] = 999,999

 Dec 4, 2019

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