A=1.2.3........100-1.2.3.....99-1.2.3.....99^2
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c) \(C=1.2+2.3+3.4+...+98.99\)
\(\Rightarrow3C=1.2\left(3-0\right)+2.3\left(4-1\right)+3.4\left(5-2\right)+...+98.99\left(100-97\right)\)
\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+98.99.100-97.98.99\)
\(=98.99.100\)
\(\Rightarrow C=\frac{98.99.100}{3}=323400\)
d) \(D=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
Chứng mình rằng stn A chia hết cho 101 với A = 1.2.3...99.100. ( 1+ 1/2 + 1/3 + ....+ 1/99 + 1/100 )
A=1•2•3•...•100
(1+1/100)+(1/2+1/99)+(1/3+1/98)+...+(1/50+1/51)
=1•2•3•100
=(101/100+101/2*99+101/3*98+...+101/50*51)
=1•2•3...100
(101/100+101/2*99+101/3*98+...+1/50*51)
Vì 101:101 => A :101
Nhớ k cho mình nha
b)B=(136/15 - 28/5 +31/5).21/24
B=(136/15-(28/5 - 31/5).21/24
B=(136/15 + 3/5).21/24
B=29/3.21/24
B=203/24
\(A=1.2.3...100-1.2.3...99-1.2.3...98.99^2\)
\(=1.2.3...99\left(100-1-99\right)\)
\(=1.2.3...99.0=0\)
Ta có: A=1.2.3.....99.100.(\(1+\dfrac{1}{2}+\dfrac{1}{3}+......+\dfrac{1}{99}+\dfrac{1}{100}\))
\(=1.2.3...100\left[\left(1+\dfrac{1}{100}\right)+\left(\dfrac{1}{2}+\dfrac{1}{99}\right)+......+\left(\dfrac{1}{50}+\dfrac{1}{51}\right)\right]\)
=>A= 1.2...100.\(\left[\dfrac{101}{100}+\dfrac{101}{2.99}+......+\dfrac{101}{50.51}\right]\)
=1.2.....100.101\(\left[\dfrac{1}{100}+\dfrac{1}{2.99}+.....+\dfrac{1}{50.51}\right]⋮101\)
Vậy A chia hết cho 101
\(A=1.2.3......100-1.2.3....99-1.2.3....99^2\)
\(=1.2.3...99\left(100-1-99\right)\)
\(=0\)
Bài làm :
Ta có :
\(A=1.2.3...100-1.2.3...99-1.2.3...99^2\)
\(A=\left(1.2.3...99\right)\left(100-1-99\right)\)
\(A=\left(1.2.3...99\right).0\)
\(A=0\)
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