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=$\frac{3^{101}-1}{4}$
