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1/2 + 2/3 + 3/4 + 4/5 + 5/6 + 6/7 + 7/8 + 8/9 + ........+ 95/96 + 96/97 + 97/98 + 98/99 + 99/100 = ?
Số các số hạng là:
(2000 - 100) : 1 + 1 = 1901
Tổng là:
(2000 + 100) x 1901 : 2 = 1996050
Đáp số : 1996050
Bài 1:
a: \(2P=2^{101}-2^{100}+2^{98}-2^{97}+...+2^3-2^2\)
=>\(3P=2^{101}-2\)
hay \(P=\dfrac{2^{101}-2}{3}\)
b: \(5Q=5^{101}-5^{100}+5^{99}-5^{98}+...+5^3-5^2+5\)
=>\(6Q=5^{101}+1\)
hay \(Q=\dfrac{5^{101}+1}{6}\)
A=-1++(-1)+..+-(1) có 50 số -1
=>A=-1x50=-50
B=(1-2-3+4)+(5-6-7+8)+...+(97-98-99+100)
B=0+0+0+..+0
B=0
C=2^100-(2^99+2^98+...+1)
C=2^100-(2^100-1)
C=1
C=\(\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
=\(\frac{1}{100}-\left(\frac{1}{2.1}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
=\(\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
=\(\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
=\(\frac{1}{100}-\frac{99}{100}\)
=\(\frac{-98}{100}=\frac{-49}{50}\)
C=1/100 -1/100.99 -1/99.98 -1/98.97-......- 1/3.2 -1/2.1
= 1/100 - (1/100.99 + 1/99.98 + 1/98.97-......+ 1/3.2 +1/2.1)
Đặt A = 1/100.99 + 1/99.98 + 1/98.97-......+ 1/3.2 +1/2.1 => C = 1/100 - A
Dễ thấy 1/2.1 = 1/1 - 1/2
1/3.2 = 1/2 - 1/3
.....................
1/99.98 = 1/98 - 1/99
1/100.99 = 1/99 - 1/100
=> cộng từng vế với vế ta
\(\frac{T}{M}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{\frac{1}{99}+\frac{2}{98}+...+\frac{98}{2}+\frac{99}{1}}\)
Xét M - 99 + 98 = \(\frac{100}{99}+\frac{100}{98}+...+\frac{100}{2}\)
\(\Leftrightarrow M-1=100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\right)\)
\(\Rightarrow M=\frac{100}{100}+100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\right)=100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(\Rightarrow\frac{T}{M}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)}=\frac{1}{100}\)
\(S=1^2+2^2+3^2+...+99^2\)
\(=1\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+...+99\left(100-1\right)\)
\(=\left(1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\right)-\left(1+2+3+...+99\right)\)
\(=\frac{99\cdot100\cdot101}{3}-\frac{99\cdot\left(99+1\right)}{2}\)
\(=333300-4950\)
\(=328350\)
\(M=1\cdot3+3\cdot5+5\cdot7+...+97\cdot99\)
\(=3+\frac{3\cdot5\cdot\left(7-1\right)+5\cdot7\cdot\left(9-3\right)+...+97\cdot99\cdot\left(101-95\right)}{6}\)
\(=3+\frac{3\cdot5\cdot7-1\cdot3\cdot5+5\cdot7\cdot9-3\cdot5\cdot7+...+97\cdot99\cdot101-95\cdot97\cdot99}{6}\)
\(=3+\frac{-\left(1\cdot3\cdot5\right)}{6}+\frac{3\cdot5\cdot7+5\cdot7\cdot9-3\cdot5\cdot7+...+97\cdot99\cdot101-95\cdot97\cdot99}{6}\)
\(=3+-\frac{15}{6}+\frac{97\cdot99\cdot101}{6}\)
\(=3+-2,5+161650,5\)
\(=161651\)