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