Tính:
A=\(\frac{2.1+1}{\left[1.\left(1+1\right)\right]^2}+\frac{2.2+1}{\left[2.\left(2+1\right)\right]^2}+...+\frac{2.99+1}{\left[99\left(99+1\right)\right]^2}\)
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
\(A=\frac{2.1+1}{\left[1\left(1+1\right)\right]^2}+\frac{2.2+1}{\left[2\left(2+1\right)\right]^2}+...+\frac{2.99+1}{\left[99\left(99+1\right)\right]^2}\)
\(=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{99^2}-\frac{1}{100^2}\)
\(=1-\frac{1}{100^2}=\frac{9999}{10000}\)
Tính B=\(\frac{2.1+1}{\left[1.\left(1+1\right)^2\right]}+\frac{2.2+1}{\left[2.\left(2+1\right)^2\right]}+\frac{2.3+1}{\left[3.\left(3+1\right)^2\right]}+...+\frac{2.99+1}{\left[99.\left(99+1\right)^2\right]}\).
tìm số nguyên a sao cho \(a^4+4\)là số nguyên tố
tính giá trị của biểu thức : \(\dfrac{2.1+1}{\left(1^2+1\right)^2}+\dfrac{2.2+1}{\left(2^2+2\right)^2}+\dfrac{2.3+1}{\left(3^2+3\right)^2}+...+\dfrac{2.2015+1}{\left(4^2+4\right)^2}+\dfrac{2.2016+1}{\left(5^2+5\right)^2}\)
\(Tìm\) \(x\)∈\(Z\)\(,\) \(biết\)\(:\)
\(a\)) \(\left(x-20\right)+\left(x-19\right)+\left(x-18\right)+...+99+100=100\)
\(b\)) \(213-x.\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}\right):\left(1-\dfrac{1}{2^{2020}}\right)=13\)
a) Quy luật là gì ??
b)
Đặt
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2020}}\\\Rightarrow2A=1+\dfrac{1}{2}+...+\dfrac{1}{2^{2019}}\\ \Rightarrow2A-A=1-\dfrac{1}{2^{2020}}\Rightarrow A=1-\dfrac{1}{2^{2020}}\)
Suy ra , phương trình trở thành :
213 -x =13
<=> x=200
f, \(x^2-x+25\)
\(=x^2-2.\dfrac{1}{2}.x+\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^2+25\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{99}{4}\)
Vì \(\left(x-\dfrac{1}{2}\right)^2\) ≥ 0 nên \(\left(x-\dfrac{1}{2}\right)^2+\dfrac{99}{4}\ge\dfrac{99}{4}\) với mọi x
Dấu "=" xảy ra ⇔ \(x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)
Vậy GTNN của đa thức là \(\dfrac{99}{4}\) tại \(x=\dfrac{1}{2}\)
tính:\(C=\left(1-\dfrac{1}{2^2}\right)\left(\dfrac{1}{3^2}-1\right)\left(1-\dfrac{1}{4^2}\right)\left(\dfrac{1}{5^2}-1\right)...\left(\dfrac{1}{99^2}-1\right)\left(1-\dfrac{1}{100^2}\right)\)
\(C=\left(\dfrac{2^2-1}{2^2}\right)\left(\dfrac{1-3^2}{3^2}\right)\left(\dfrac{4^2-1}{4^2}\right)...\left(\dfrac{1-99^2}{100^2}\right)\left(\dfrac{100^2-1}{99^2}\right)=\left(\dfrac{1.3}{2^2}\right)\left(\dfrac{-2.4}{3^2}\right)\left(\dfrac{3.5}{4^2}\right)...\left(\dfrac{-98.100}{99^2}\right)\left(\dfrac{99.101}{100^2}\right)=-\dfrac{101}{200}\)
\(B=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{98}+\left(\dfrac{1}{2}\right)^{99}\)
So sánh B với 1
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
a, \(\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{2^2}\right)\left(1+\dfrac{1}{2^3}\right)\left(1+\dfrac{1}{2^4}\right)....\left(1+\dfrac{1}{2^{50}}\right)< 3\)
b, \(\dfrac{1}{2}-\dfrac{1}{2^2}+.........+\dfrac{1}{2^{99}}-\dfrac{1}{2^{100}}< \dfrac{1}{3}\)
\(\left(1+\dfrac{1}{2}\right)+\left(1+\dfrac{1}{2^2}\right)+...+\left(1+\dfrac{1}{2^{50}}\right)\)
= \(\left(1+1+1+...+1\right)+\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{50}}\right)\)(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-\(\dfrac{1}{2^{50}}>3\)
B= (\(1+\dfrac{1}{2}\))\(\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{4}\right)...\left(1+\dfrac{1}{99}\right)\)
Ta có: \(B=\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{4}\right)\cdot...\cdot\left(1+\dfrac{1}{99}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{100}{99}\)
\(=\dfrac{100}{2}=50\)
