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# ​ Method of difference

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Thank you

Dec 13, 2018

#1
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Method of difference

Formula:

$$\begin{array}{|rcll|} \hline \displaystyle \sum \limits_{r=1}^{n}r^2 &=& \dfrac16n(n+1)(2n+1) \\\\ \displaystyle \sum \limits_{r=1}^{2n}r^3 &=& \dfrac1{2^2} (2n)^2(1+2n)^2 =n^2(1+2n)^2 \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline &&\mathbf{ \displaystyle \sum \limits_{r=1}^{n}(6r-3)^2 } \\\\ &=& \displaystyle \sum \limits_{r=1}^{n}(36r^2-36r+9) \\\\ &=& \displaystyle 36\sum \limits_{r=1}^{n}r^2- 36 \sum \limits_{r=1}^{n}r+ 9\sum \limits_{r=1}^{n}1 \\\\ &=& 36\dfrac16n(n+1)(2n+1) -36\dfrac{(n+1)}{2}n +9n \\\\ &=& 6n(n+1)(2n+1) -18n(n+1) +9n \\\\ &=& 6n(n+1)(2n+1) -18n^2-18n+9n \\\\ &=& 6n(n+1)(2n+1) -18n^2-9n \\\\ &=& 3n\Big( 2(n+1)(2n+1) -6n-3 \Big) \\\\ &=& 3n\Big( 2(n+1)(2n+1) -3(2n+1) \Big) \\\\ &=& 3n(2n+1)\Big( 2(n+1) -3 \Big) \\\\ &=& 3n(2n+1)(2n-1) \\\\ &\mathbf{=}& \mathbf{3n(4n^2-1)} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline && \mathbf{\displaystyle \sum \limits_{r=1}^{2n}r^3 - \sum \limits_{r=1}^{n}(6r-3)^2 } \\\\ &=& n^2(1+2n)^2 - 3n(4n^2-1) \\ &=& n^2(1+2n)^2 - 3n(2n-1)(2n+1) \\ &=& n(1+2n)\Big(n(1+2n) - 3(2n-1)\Big) \\ &=& n(1+2n)(n+2n^2 - 6n +3 ) \\ &=& n(1+2n)(2n^2 -5n +3 ) \\ &\mathbf{=}& \mathbf{ n(1+2n)(2n-3)(n-1) } \\ \hline \end{array}$$

Dec 13, 2018