$\begin{align}
& B=\left( \frac{1}{3}+\frac{1}{{{3}^{2}}}+\frac{1}{{{3}^{3}}}+\frac{1}{{{3}^{4}}} \right){{.3}^{5}}+\left( \frac{1}{{{3}^{5}}}+\frac{1}{{{3}^{6}}}+\frac{1}{{{3}^{7}}}+\frac{1}{{{3}^{8}}} \right){{.3}^{9}}+...+\left( \frac{1}{{{3}^{97}}}+\frac{1}{{{3}^{98}}}+\frac{1}{{{3}^{99}}}+\frac{1}{{{3}^{100}}} \right){{.3}^{101}} \\
& B=({{3}^{4}}+{{3}^{3}}+{{3}^{2}}+3)+({{3}^{4}}+{{3}^{3}}+{{3}^{2}}+3)+...+({{3}^{4}}+{{3}^{3}}+{{3}^{2}}+3) \\
\end{align}$
Có tất cả: (101 – 5) : 4 + 1 = 25 (cặp)
B = 25 $\times $ $({{3}^{4}}+{{3}^{3}}+{{3}^{2}}+3)$
A = ${{3}^{4}}+{{3}^{3}}+{{3}^{2}}+3$
A $\times $ 3 = $({{3}^{4}}+{{3}^{3}}+{{3}^{2}}+3)\times 3={{3}^{5}}+{{3}^{4}}+{{3}^{3}}+{{3}^{2}}$
A $\times $ 3 – A = ${{3}^{5}}-3$= 240
A = 120
B = 3000

A=120
B=3000 nha
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