chứng minh 5^ ( n+3) – 3^(n+3) +5^(n+2)- 3^(n+1) chia hết cho 60 b, cho C = 1/4 + 2/4^2 + 3/4^3 +…….. +2017/4^2017 chúng ming rằng C < 1/2 giú

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1. Giải thích các bước giải:

   a.Ta có:

   $A=5^{n+3}-3^{n+3}+5^{n+2}-3^{n+1}$
   $\to A=(5^{n+3}+5^{n+2})-(3^{n+3}+3^{n+1})$ 

   $\to A=5^{n+2}(5+1)-3^{n+1}(3^2+1)$ 

   $\to A=5^{n+2}\cdot 6-3^{n+1}\cdot 10$ 

   $\to A=5^{n+1}\cdot 5\cdot 6-3^{n}\cdot 3\cdot 10$ 

   $\to A=5^{n+1}\cdot 30-3^{n}\cdot 30$ 

   $\to A=30(5^{n+1}-3^{n})$ 

   Mà $5^{n+1},3^{n}$ lẻ

   $\to 5^{n+1}-3^{n}$ chẵn

   $\to 5^{n+1}-3^{n}\quad\vdots\quad 2$

   $\to 30(5^{n+1}-3^{n})\quad\vdots\quad 60$

   $\to A\quad\vdots\quad 60$

   b.Ta có:

   $C=\dfrac14+\dfrac2{4^2}+\dfrac3{4^3}+…+\dfrac{2017}{4^{2017}}$

   $\to 4C=1+\dfrac2{4}+\dfrac3{4^2}+…+\dfrac{2017}{4^{2016}}$

   $\to 4C-C=1+\dfrac14+\dfrac1{4^2}+…+\dfrac{1}{4^{2016}}-\dfrac{2017}{4^{2017}}$

   $\to 3C=1+\dfrac14+\dfrac1{4^2}+…+\dfrac{1}{4^{2016}}-\dfrac{2017}{4^{2017}}$

   Mà $B=1+\dfrac14+\dfrac1{4^2}+…+\dfrac{1}{4^{2016}}$

   $\to 4B=4+1+\dfrac14+…+\dfrac1{4^{2015}}$

   $\to 4B-B=4-\dfrac{1}{4^{2016}}$

   $\to 3B=4-\dfrac{1}{4^{2016}}$

   $\to B=\dfrac43-\dfrac{1}{3\cdot 4^{2016}}$

   $\to 3C=(1+\dfrac14+\dfrac1{4^2}+…+\dfrac{1}{4^{2016}})-\dfrac{2017}{4^{2017}}$

   $\to 3C=B-\dfrac{2017}{4^{2017}}$

   $\to 3C=\dfrac43-\dfrac{1}{3\cdot 4^{2016}}-\dfrac{2017}{4^{2017}}$

   $\to 3C<\dfrac43$

   $\to C<\dfrac49<\dfrac48=\dfrac12$

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