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a)
Áp dụng công thức (a - b).(a+ b) = a.(a+ b) - b.(a+ b) = a2 + ab - ab - b2 = a2 - b2
Ta có
\(M=100^2-99^2+98^2-97^2+...+2^2-1^2\)
M = (100 - 99)(100 + 99) + (98 - 97).(98 + 97) + ...+ (2 - 1)(2+1)
= 100 + 99 + 98 + 97 + ...+ 2 + 1
= (1+100).100 : 2
= 5050
b)
N = (202 - 192 ) + (182 - 172 ) + ...+ (42 - 32 ) + (22 - 12 )
= (20 - 19).(20 + 19) + (18 - 17)(18 + 17) +...+ (4 -3)(4 +3) + (2-1)(2+1) = 39 + 35 + ...+ 7 + 3
N = (39 + 3).10 : 2 = 210
$B=\left(\dfrac1{2^2}-1\right)\left(\dfrac1{3^2}-1\right)\cdots\left(\dfrac1{99^2}-1\right)$
$=\left(-\dfrac{2^2-1}{2^2}\right)\left(-\dfrac{3^2-1}{3^2}\right)\cdots\left(-\dfrac{99^2-1}{99^2}\right)$
$=(-1)^{98}\prod_{k=2}^{99}\dfrac{(k-1)(k+1)}{k^2}$
$=\left(\prod_{k=2}^{99}\dfrac{k-1}{k}\right)\left(\prod_{k=2}^{99}\dfrac{k+1}{k}\right)$
$=\left(\dfrac12\cdot\dfrac23\cdot\dfrac34\cdots\dfrac{98}{99}\right)\left(\dfrac32\cdot\dfrac43\cdot\dfrac54\cdots\dfrac{100}{99}\right)$
$=\dfrac1{99}\cdot\dfrac{100}{2}$
$=\dfrac{50}{99}.$
$\textbf{a)}$
$A=\left(1-\dfrac12\right)\left(1-\dfrac13\right)\left(1-\dfrac14\right)\cdots\left(1-\dfrac1n\right)$
$=\dfrac12\cdot\dfrac23\cdot\dfrac34\cdots\dfrac{n-1}{n}$
$=\dfrac12\cdot\dfrac23\cdot\dfrac34\cdots\dfrac{n-1}{n}$
$=\dfrac1n.$
$\textbf{b)}$
$B=\left(1-\dfrac1{2^2}\right)\left(1-\dfrac1{3^2}\right)\cdots\left(1-\dfrac1{n^2}\right)$
$=\dfrac{(2-1)(2+1)}{2^2}\cdot\dfrac{(3-1)(3+1)}{3^2}\cdots\dfrac{(n-1)(n+1)}{n^2}$
$=\left(\dfrac12\cdot\dfrac23\cdot\dfrac34\cdots\dfrac{n-1}{n}\right)\left(\dfrac32\cdot\dfrac43\cdot\dfrac54\cdots\dfrac{n+1}{n}\right)$
$=\dfrac1n\cdot\dfrac{n+1}{2}$
$=\dfrac{n+1}{2n}.$
Bài 1 :\(a,=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{100^2}{99.101}\)
\(=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4...101}\)
\(=100.\frac{2}{101}=\frac{200}{101}\)
\(A=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-1\frac{15}{17}+\frac{2}{3}=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-\frac{64}{34}+\frac{14}{21}=\left(\frac{15}{34}+\frac{9}{34}-\frac{64}{34}\right)+\left(\frac{7}{21}+\frac{14}{21}\right)=\frac{30}{34}+\frac{21}{21}=\frac{15}{17}+1=\frac{32}{17}\)