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\(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-...-\frac{100}{3^{100}}\)
\(\Rightarrow3A=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow3A+A=1+\left(\frac{1}{3}-\frac{2}{3}\right)+\left(\frac{-2}{3^2}+\frac{3}{3^2}\right)+\left(\frac{3}{3^3}-\frac{4}{3^3}\right)+...+\left(\frac{-98}{3^{98}}+\frac{99}{3^{98}}\right)+\left(\frac{99}{3^{99}}-\frac{100}{3^{99}}\right)-\frac{100}{3^{100}}\)
\(\Rightarrow4A=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow3.4A=3-1+\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow3.4A+4A=3+\left(1-1\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{3^2}-\frac{1}{3^2}\right)+...+\left(\frac{1}{3^{98}}-\frac{1}{3^{98}}\right)-\frac{101}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow16A=3-\frac{99}{3^{99}}-\frac{100}{3^{100}}< 3\Rightarrow A< \frac{3}{16}< \frac{3}{4}\)
$A=\dfrac12-\dfrac{2}{2^2}+\dfrac{3}{2^3}-\dfrac{4}{2^4}+\cdots+\dfrac{99}{2^{99}}-\dfrac{100}{2^{100}}$
Nhóm từng 2 số:
$A=\left(\dfrac12-\dfrac{2}{2^2}\right)+\left(\dfrac3{2^3}-\dfrac4{2^4}\right)+\cdots+\left(\dfrac{99}{2^{99}}-\dfrac{100}{2^{100}}\right)$
$=0+\dfrac18+\dfrac{2}{32}+\dfrac3{128}+\cdots+\dfrac{49}{2^{99}}$
$=\sum_{k=1}^{50}\dfrac{k-1}{2^{2k-1}}$
Ta có: $\dfrac{k-1}{2^{2k-1}}=\dfrac{2(k-1)}{4^k}$
Mà: $\sum_{k=1}^{\infty}\dfrac{k-1}{4^k}=\dfrac{1}{9}$
Nên: $A<2\cdot\dfrac19$ $=\dfrac29$
Vậy: $A<\dfrac29$
b)$4=1\cdot4,\quad28=4\cdot7,\quad70=7\cdot10,\ldots$
tức là: $E=\dfrac3{1\cdot4}+\dfrac3{4\cdot7}+\dfrac3{7\cdot10}+\cdots+\dfrac3{n(n+3)}$
Với $n=1,4,7,\ldots$.
Ta có: $\dfrac3{n(n+3)}=\dfrac1n-\dfrac1{n+3}$
Do đó: $E=\left(1-\dfrac14\right)+\left(\dfrac14-\dfrac17\right)+\left(\dfrac17-\dfrac1{10}\right)+\cdots+\left(\dfrac1n-\dfrac1{n+3}\right)$
$=1-\dfrac1{n+3}$
Vì: $\dfrac1{n+3}>0$ nên: $1-\dfrac1{n+3}<1$