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Question-109640




Question Number 109640 by hhryhrry2 last updated on 24/Aug/20
Commented by kaivan.ahmadi last updated on 24/Aug/20
a.  Σ_2 ^6 (g^2 +2g)−Σ_2 ^6 (3(g+3)−5)=Σ_2 ^6 (g^2 −g−4)=  Σ_3 ^7 ((g−1)^2 −(g−1)−4)=Σ_3 ^7 (g^2 −3g−4)=  Σ_3 ^7 (g−4)(g+1)
$${a}. \\ $$$$\underset{\mathrm{2}} {\overset{\mathrm{6}} {\sum}}\left({g}^{\mathrm{2}} +\mathrm{2}{g}\right)−\underset{\mathrm{2}} {\overset{\mathrm{6}} {\sum}}\left(\mathrm{3}\left({g}+\mathrm{3}\right)−\mathrm{5}\right)=\underset{\mathrm{2}} {\overset{\mathrm{6}} {\sum}}\left({g}^{\mathrm{2}} −{g}−\mathrm{4}\right)= \\ $$$$\underset{\mathrm{3}} {\overset{\mathrm{7}} {\sum}}\left(\left({g}−\mathrm{1}\right)^{\mathrm{2}} −\left({g}−\mathrm{1}\right)−\mathrm{4}\right)=\underset{\mathrm{3}} {\overset{\mathrm{7}} {\sum}}\left({g}^{\mathrm{2}} −\mathrm{3}{g}−\mathrm{4}\right)= \\ $$$$\underset{\mathrm{3}} {\overset{\mathrm{7}} {\sum}}\left({g}−\mathrm{4}\right)\left({g}+\mathrm{1}\right) \\ $$
Commented by kaivan.ahmadi last updated on 24/Aug/20
b.  Σ_5 ^9 (2b−5)^2 =Σ_1 ^5 (2(b+4)−5)^2 =Σ_1 ^5 (2b−3)^2 =  Σ_1 ^5 (4b^2 −12b+9)=4Σ_1 ^5 b^2 −3Σ_1 ^5 (4b−3)
$${b}. \\ $$$$\underset{\mathrm{5}} {\overset{\mathrm{9}} {\sum}}\left(\mathrm{2}{b}−\mathrm{5}\right)^{\mathrm{2}} =\underset{\mathrm{1}} {\overset{\mathrm{5}} {\sum}}\left(\mathrm{2}\left({b}+\mathrm{4}\right)−\mathrm{5}\right)^{\mathrm{2}} =\underset{\mathrm{1}} {\overset{\mathrm{5}} {\sum}}\left(\mathrm{2}{b}−\mathrm{3}\right)^{\mathrm{2}} = \\ $$$$\underset{\mathrm{1}} {\overset{\mathrm{5}} {\sum}}\left(\mathrm{4}{b}^{\mathrm{2}} −\mathrm{12}{b}+\mathrm{9}\right)=\mathrm{4}\underset{\mathrm{1}} {\overset{\mathrm{5}} {\sum}}{b}^{\mathrm{2}} −\mathrm{3}\underset{\mathrm{1}} {\overset{\mathrm{5}} {\sum}}\left(\mathrm{4}{b}−\mathrm{3}\right) \\ $$$$ \\ $$

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