21=45

đặt A=1/2+(1/2)^2+(1/2)^3+...+(1/2)^98+(1/2)^99+(1/2)^99

=>A=1/2+1²/2²+1³/2³+...+1⁹⁸/2⁹⁸+1⁹⁹/2⁹⁹+1⁹⁹/2⁹⁹

=>A=1/2+1/2²+1/2³+...+1/2⁹⁸+1/2⁹⁹+1/2⁹⁹

=>2A-1/2⁹⁹=1+1/2+1/2²+...+1/2⁹⁸

=>(2A-1/2⁹⁹)-(A-1/2⁹⁹)=(1+1/2+1/2²+...+1/2⁹⁸)-(1/2+1/2²+1/2³+...+1/2⁹⁸+1/2⁹⁹)

=>(2A-1/2⁹⁹)-(A-1/2⁹⁹)=1-1/2⁹⁹

=>A=1-1/2⁹⁹ +1/2⁹⁹=1

vậy A=1

chắc thế

cái phân số cuối sai thì phải 

Nhân 1/2B lên , ta được B=1

Nhân 1/2B lên, ta được:

1/2B=(1/2)^2+(1/2)^3+...+(1/2)^100

lấy B-1/2B, ta được: 1/2B=1/2-(1/2)^100

                           =>B= (1/2-(1/2)^100)/2

Đặt \(A=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{3}\right)^3+...+\left(\frac{1}{2}\right)^{99}\)

\(\Rightarrow A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)

\(\Rightarrow2A-A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{99}}\)

\(\Rightarrow A=1-\frac{1}{2^{99}}=\frac{2^{99}-1}{2^{99}}\)

\(A=\frac{1}{2}+\frac{1}{2^2}+.............+\frac{1}{2^{99}}\)

\(\Leftrightarrow2A=1+\frac{1}{2}+...........+\frac{1}{2^{98}}\)

\(\Leftrightarrow2A-A=\left(1+\frac{1}{2}+.......+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+.......+\frac{1}{2^{99}}\right)\)

\(\Leftrightarrow A=1-\frac{1}{2^{99}}\)

\(\Leftrightarrow2^{99}.A=2^{99}-1\left(đpcm\right)\)

$B=\left(\dfrac1{2^2}-1\right)\left(\dfrac1{3^2}-1\right)\cdots\left(\dfrac1{99^2}-1\right)$

$=\left(-\dfrac{2^2-1}{2^2}\right)\left(-\dfrac{3^2-1}{3^2}\right)\cdots\left(-\dfrac{99^2-1}{99^2}\right)$

$=(-1)^{98}\prod_{k=2}^{99}\dfrac{(k-1)(k+1)}{k^2}$

$=\left(\prod_{k=2}^{99}\dfrac{k-1}{k}\right)\left(\prod_{k=2}^{99}\dfrac{k+1}{k}\right)$

$=\left(\dfrac12\cdot\dfrac23\cdot\dfrac34\cdots\dfrac{98}{99}\right)\left(\dfrac32\cdot\dfrac43\cdot\dfrac54\cdots\dfrac{100}{99}\right)$

$=\dfrac1{99}\cdot\dfrac{100}{2}$

$=\dfrac{50}{99}.$