Tính N=3^2022-2.(3^2021+3^2020+…+3^1+3^0)

1 bình luận về “Tính N=3^2022-2.(3^2021+3^2020+…+3^1+3^0)”

1. Giải đáp: $N=1$

    

   Lời giải và giải thích chi tiết:

   $N={{3}^{2022}}-2.\left( {{3}^{2021}}+{{3}^{2020}}+…+{{3}^{1}}+{{3}^{0}} \right)$

   Đặt $A={{3}^{0}}+{{3}^{1}}+…+{{3}^{2020}}+{{3}^{2021}}$

   $\Rightarrow 3A={{3}^{1}}+{{3}^{2}}+…+{{3}^{2021}}+{{3}^{2022}}$

   $\Rightarrow 3A-A={{3}^{2022}}-{{3}^{0}}$

   $\Rightarrow 2A={{3}^{2022}}-1$

   $\Rightarrow A=\dfrac{{{3}^{2022}}-1}{2}$

   Vậy $N={{3}^{2022}}-2\cdot \dfrac{{{3}^{2022}}-1}{2}$

   $N={{3}^{2022}}-\left( {{3}^{2022}}-1 \right)$

   $N={{3}^{2022}}-{{3}^{2022}}+1$

   $N=1$

   Trả lời