Cho A = (1/2^2 – 1) . (1/3^2 – 1) . (1/4^2 – 1) … (1/100^2 -1). So sánh A với (-1/2) Rút gọn A = 2^100 – 2^99 + 2^98 – 2^97 +….+ 2^2 – 2 Rút gọn B

0 bình luận về “Cho A = (1/2^2 – 1) . (1/3^2 – 1) . (1/4^2 – 1) … (1/100^2 -1). So sánh A với (-1/2) Rút gọn A = 2^100 – 2^99 + 2^98 – 2^97 +….+ 2^2 – 2 Rút gọn B”

1. Giải thích các bước giải:

   `A=(1/2^2-1).(1/3^2-1).(1/4^2-1)…(1/100^2-1)`

   `=(-3)/4.(-8)/9.(-15)/16….(-99)/100`

   `=-(3/4. 8/9. 15/16 … 99/100)`

   `=-((1.3)/(2.2). (2.4)/(3.3).(3.5)/(4.4)…(9.11)/(10.10))`

   `=-((1.2.3…9).(3.4.5…11))/((2.3.4…10).(2.3.4…10))`

   `=-(1.11)/(10.2)=(-11)/20< (-1)/20=(-1)/2`

   .

   `A=2^100-2^99+2^98-2^97+…+2^2-2`

   `=> 2A=2^101 – 2^100+2^99-2^98+…+2^3-2^2`

   `=>2A + A=2^101-2`

   `=>3A=2^101-2`

   `=>A=(3^201-2)/3`

   .

   `B=3^100-3^99+3^98-3^97+…+3^2-3+1`

   `=>3B = 3^101-3^100+3^99-3^98+…+3^3-3^2+3`

   `=>3B+B=3^101+1`

   `=>4B=3^101+1`

   `=>B=(3^101+1)/4`

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2. Đáp án:

   Quốc Đẹp Trai

   Giải thích các bước giải:

    A=A=$\frac{1}{2^{2}-1}$ .$\frac{1}{3^{2}-1}$.$\frac{1}{4^{2}-1}$….$\frac{1}{100^{2}-1}$

   A=$\frac{1}{8}$. $\frac{1}{15}$.$\frac{1}{24}$….$\frac{1}{9999}$

   A=$\frac{1}{2.4}$.$\frac{1}{3.5}$.$\frac{1}{4.6}$.$\frac{1}{99.101}$

   A=$\frac{1}{2}$ .($\frac{1}{1}$- $\frac{1}{3}$) .$\frac{1}{2}$ .($\frac{1}{2}$- $\frac{1}{4}$) …..$\frac{1}{2}$ .($\frac{1}{99}$- $\frac{1}{101}$)

   Do A>0 =>A>$\frac{-1}{2}$ 

   1)A=2^100 – 2^99 + 2^98 – 2^97 +….+ 2^2 – 2

   =>2A=2^101 – 2^100 + 2^99 – 2^98 +….+ 2^3 – 2^2

   =>3A=(2^101 – 2^100 + 2^99 – 2^98 +….+ 2^3 – 2^2)+(2^100 – 2^99 + 2^98 – 2^97 +….+ 2^2 – 2)

   3A=2^101-2

   =>A=$\frac{2^{101-2}}{3}$ 

   2) B = 3^100 – 3^99 + 3^98 – 3^97 +….+ 3^2 – 3 + 1

   =>3B= 3(3^100 – 3^99 + 3^98 – 3^97 +….+ 3^2 – 3 + 1)

   3B= 3^101 – 3^100 + 3^99 – 3^98 +….+ 3^3 – 3^2 +3

   =>3B+B=4B=(3^101 – 3^100 + 3^99 – 3^98 +….+ 3^3 – 3^2 +3)+( 3^100 – 3^99 + 3^98 – 3^97 +….+ 3^2 – 3 + 1)

   4B=3^101-1

   =>B=$\frac{3^{101}-1}{4}$ 

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