[tex]a=1+7+7^2+\ldots+7^{2011}\\
\text{Tinem cont de faptul ca in total sunt 2012 termeni,deci ii putem }\\
\text{grupa cate 2.}\\
a=(1+7)+7^2(1+7)+\ldots+7^{2010}(1+7)\\
a=(1+7)(1+7^2+7^4+\ldots+7^{2010})\\
a=8\cdot (1+7^2+7^4+\ldots 7^{2010})\ \vdots \ 4\\
b=3+3^2+3^3+\ldots +3^{27}\\
\text{Procedam la fel ca la exercitiul anterior.}\\
b=3(1+3+3^2)+3^4(1+3+3^2)+\ldots +3^{25}(1+3+3^2)\\
b=(1+3+3^2)(3+3^4+\ldots +3^{25})\\
b=13(3+3^4+\ldots +3^{25})\ \vdots\ 13[/tex]