cho A=3+3^2+3^3+…+3^100.tìm n sao cho 2×A+3=3^n

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1. Ta có: A = 3 + $3^{2}$ + $3^{3}$ +…+$3^{100}$

   => 3A = $3^{2}$ + $3^{3}$ + $3^{4}$ +…+$3^{101}$

   => 3A – A = $3^{101}$ – 3

   Mà 2A + 3 = $3^{n}$

   hay $3^{101}$ – 3 + 3 = $3^{n}$

   => $3^{101}$ = $3^{n}$

   => n = 101

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2. $A=3+3^2+…+3^{100}$

   $⇒3A=3^2+3^3+…+3^{101}$

   $⇒3A-A=-1+3^{101}$

   $⇒2A=3^{101}-3$

   $⇒A=$$\frac{3^{101}-3}{2}$

   $⇒2A+3=$$\frac{2.(3^{101}-3)}{2}+3$ 

   $=3^{101}=3^n$

   $⇒n=101$

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