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Ta có:
\(\frac{2n+1}{\left[n\left(n+1\right)\right]^2}=\frac{n+n+1}{n^2\left(n+1\right)^2}=\frac{1}{n\left(n+1\right)^2}+\frac{1}{n^2\left(n+1\right)}\)
\(=\frac{1}{n\left(n+1\right)}.\left(\frac{1}{n}+\frac{1}{n+1}\right)=\left(\frac{1}{n}-\frac{1}{n+1}\right).\left(\frac{1}{n}+\frac{1}{n+1}\right)\)
\(=\frac{1}{n^2}-\frac{1}{\left(n+1\right)^2}\)
Áp dụng vào bài toán ta được
\(A=\frac{2.1+1}{\left[1\left(1+1\right)\right]^2}+\frac{2.2+1}{\left[2\left(2+1\right)\right]^2}+...+\frac{2.99+1}{\left[99\left(99+1\right)\right]^2}\)
\(=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{99^2}-\frac{1}{100^2}\)
\(=1-\frac{1}{100^2}=\frac{9999}{10000}\)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)
b, \(\left(1-\dfrac{1}{100}\right)\left(1-\dfrac{1}{99}\right)...\left(1-\dfrac{1}{2}\right)=\dfrac{99.98...1}{100.99...2}=\dfrac{1}{100}\)
ta có : \(B=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{99}\)
\(\Rightarrow\dfrac{1}{2}B=\dfrac{1}{2}\left(\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{99}\right)=\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{100}\) \(\Rightarrow B-\dfrac{1}{2}B=\dfrac{1}{2}B=\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^{100}\)
\(\Rightarrow B=2.\dfrac{1}{2}B=1\left(\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^{100}\right)=1-\left(\dfrac{1}{2}\right)^{99}< 1\)
vậy \(B< 1\)
2B= 1+ 1/2+ (1/2)2+ ....+(1/2)98
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B= 1/2+ (1/2)2+ ....+(1/2)99
B= 1- (1/2)99 <1
=>B <1
\(\left(1+\dfrac{1}{2}\right)+\left(1+\dfrac{1}{2^2}\right)+...+\left(1+\dfrac{1}{2^{50}}\right)\)
= \(\left(1+1+1+...+1\right)+\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{50}}\right)\)(50 số 1 )
= \(50+\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{50}}\right)\)
A =\(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{50}}\)
⇒ 2A = \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{49}}\)
⇒ 2A - A =\(1-\dfrac{1}{2^{50}}\)
=50+1-\(\dfrac{1}{2^{50}}\)=51-\(\dfrac{1}{2^{50}}>3\)
Ta có: \(B=\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{4}\right)\cdot...\cdot\left(1+\dfrac{1}{99}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{100}{99}\)
\(=\dfrac{100}{2}=50\)
b) Giải:
ĐK: \(a\ne-b\)
Ta có:
\(3a^2+b^2=4ab\)
\(\Leftrightarrow4a^2-4ab+b^2-a^2=0\)
\(\Leftrightarrow\left(2a-b\right)^2-a^2=0\)
\(\Leftrightarrow\left(3a-b\right)\left(a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3a-b=0\\a-b=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}a=\dfrac{b}{3}\\a=b\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a=\dfrac{b}{3}\Leftrightarrow P=\dfrac{\dfrac{b}{3}-b}{\dfrac{b}{3}+b}=\dfrac{-1}{2}\\a=b\Leftrightarrow P=\dfrac{a-a}{a+a}=\dfrac{0}{2a}=0\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}P=\dfrac{-1}{2}\\P=0\end{matrix}\right.\)