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\(3S=1+\dfrac{1}{3}+...+\dfrac{1}{3^{99}}\)
=>2S=1-1/3^100
=>S=1/2-1/2*3^100<1/2
1/2^2+1/3^2+...+1/50^2<1/1*2+1/2*3*+...+1/49*50
=1/1-1/2+1/2-1/3+...+1/49-1/50<1
=>S<1+1=2
S = 1/3 + 1/3^2 + 1/3^3 + 1/3^4 + ... + 1/3^99 + 1/3^100
3S = 1 +1/3 +1/3^2 +1/3^3 + ... + 1/3^98 +1/3^99
3S - S = ( 1 + 1/3 + 1/3^2 +1/^3 + ... + 1/3^98 +1/3^99 ) - ( 1/3 + 1/3^2 + 1/3^3 + 1/3^4 +... + 1/3^99 + 1/3^100 )
2S = 1 - 1/3^100
S = (1 - 1/3^100). 1/2
\(P=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)
\(P=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+1\)
\(P=\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}+\dfrac{50}{50}\)
\(P=50\left(\dfrac{1}{2}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)\)
\(\dfrac{S}{P}=\dfrac{\dfrac{1}{2}+...+\dfrac{1}{49}+\dfrac{1}{50}}{50\left(\dfrac{1}{2}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)}=\dfrac{1}{50}\)
a.
$S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}$
$2S=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{49}}$
$\Rightarrow 2S-S=1-\frac{1}{2^{50}}$
$\Rightarrow S=1-\frac{1}{2^{50}}$
b.
$S=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{20}}$
$3S=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{19}}$
$\Rightarrow 3S-S=1-\frac{1}{3^{20}}$
$\Rightarrow 2A=1-\frac{1}{3^{20}}$
$\Rightarrow A=\frac{1}{2}-\frac{1}{2.3^{20}}$
@Trần Bảo Việt, bạn không trả lời linh tinh nhé!