Câu hỏi của Aug.21

Question

Bài 1: a,   Rút gọn $A=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}$b, Tìm số tự nhiên thỏa mãn điều kiện: \(2.2^2+3.2^3+4.2^4+...+\left(n-1\right)....

Aug.21 · Accepted Answer

$a,A=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-..-\frac{1}{3.2}-\frac{1}{2.1}$$A=\frac{1}{100}-\left(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\right)$$A=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)$$A=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)$$A=\frac{1}{100}-\left(1-\frac{1}{100}\right)$$A=\frac{1}{100}-1+\frac{1}{100}$$A=\frac{2}{100}-1$$A=\frac{1}{50}-1$$A=\frac{-49}{50}$Câu trả lời: $a,A=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-..-\frac{1}{3.2}-\frac{1}{2.1}$$A=\frac{1}{100}-\left(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\right)$$A=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)$$A=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)$$A=\frac{1}{100}-\left(1-\frac{1}{100}\right)$$A=\frac{1}{100}-1+\frac{1}{100}$$A=\frac{2}{100}-1$$A=\frac{1}{50}-1$$A=\frac{-49}{50}$

Aug.21 · Answer

b,$2.2^2+3.2^3+4.2^4+...+\left(n-1\right).2^{n-1}+n.2^n=2^{n+34}$        (1)Đặt $B=2.2^2+3.2^3+4.2^4+...+\left(n-1\right).2^{n-1}+n.2^n$$\Rightarrow2B=2.\left(2.2^2+3.2^3+4.2^4+...+\left(n-1\right).2^{n-1}+n.2^n\right)$             $=2.2^3+3.2^4+4.2^5+...+\left(n-1\right).2^n+n.2^{n+1}$$2B-B=\left(2.2^3+3.2^4+4.2^5+..+\left(n-1\right).2^n+n.2^{n+1}\right)$                 $=(2.2^2+3.2^3+4.2^4+...+\left(n-1\right).2^{n-1}+n.2^n)$             $B=-2^3-2^4-2^5-...-2^{n+1}-2.2^2$                 $=-\left(2^3+2^4+2^5+...+2^n\right)+n.2^{n+1}-2^3$Đặt $C=2^3+2^4+2^5+2^n$$\Rightarrow2C=2.(2^3+2^4+2^5+...+2^n)$         $C=2^4+2^5+2^6+...+2^{n+1}$$2C-C=\left(2^4+2^5+2^6+...+2^{n+1}\right)-\left(2^3+2^4+2^5+...+2^n\right)$$C=2^{n+1}-2^3$Khi đó :  $B=-(2^{n+1}-2^3)+n.2^{n+1}-2^3$                  $=-2^{n+1}+2^3+n.2^{n+1}-2^3$                   =$=-2^{n+1}+n.2^{n+1}=\left(n-1\right).2^{n-1}$Vậy từ (1) ta có:$\left(n-1\right),2^{n+1}=2^{n+34}$                           $2^{n+34}-\left(n-1\right).2^{n+1}=0$                          $2^{n+1}.[2^{33}-\left(n-1\right)]=0$Do đó $2^{33}-n+1=0$( Vì $2^{n+1}\ne0$với mọi $n$)$n=2^{33}+1$Vậy $n=2^{33}+1$Câu trả lời: b,$2.2^2+3.2^3+4.2^4+...+\left(n-1\right).2^{n-1}+n.2^n=2^{n+34}$        (1)Đặt $B=2.2^2+3.2^3+4.2^4+...+\left(n-1\right).2^{n-1}+n.2^n$$\Rightarrow2B=2.\left(2.2^2+3.2^3+4.2^4+...+\left(n-1\right).2^{n-1}+n.2^n\right)$             $=2.2^3+3.2^4+4.2^5+...+\left(n-1\right).2^n+n.2^{n+1}$$2B-B=\left(2.2^3+3.2^4+4.2^5+..+\left(n-1\right).2^n+n.2^{n+1}\right)$                 $=(2.2^2+3.2^3+4.2^4+...+\left(n-1\right).2^{n-1}+n.2^n)$             $B=-2^3-2^4-2^5-...-2^{n+1}-2.2^2$                 $=-\left(2^3+2^4+2^5+...+2^n\right)+n.2^{n+1}-2^3$Đặt $C=2^3+2^4+2^5+2^n$$\Rightarrow2C=2.(2^3+2^4+2^5+...+2^n)$         $C=2^4+2^5+2^6+...+2^{n+1}$$2C-C=\left(2^4+2^5+2^6+...+2^{n+1}\right)-\left(2^3+2^4+2^5+...+2^n\right)$$C=2^{n+1}-2^3$Khi đó :  $B=-(2^{n+1}-2^3)+n.2^{n+1}-2^3$                  $=-2^{n+1}+2^3+n.2^{n+1}-2^3$                   =$=-2^{n+1}+n.2^{n+1}=\left(n-1\right).2^{n-1}$Vậy từ (1) ta có:$\left(n-1\right),2^{n+1}=2^{n+34}$                           $2^{n+34}-\left(n-1\right).2^{n+1}=0$                          $2^{n+1}.[2^{33}-\left(n-1\right)]=0$Do đó $2^{33}-n+1=0$( Vì $2^{n+1}\ne0$với mọi $n$)$n=2^{33}+1$Vậy $n=2^{33}+1$

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