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\(=\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+100\right)}{100.1+99.2+98.3+...+2.99+1.100}\)
\(=\frac{1.100+2.99+3.98+...+99.2+100.1}{100.1+99.2+98.3+...+2.99+1.100}\)
\(=1\)
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Thoe công thức dãy óố viết theo qui luật, đúng 100%. bạn chọn đúng đi !
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\(E=1.1+2.2+3.3+4.4+...+99.99\)
\(\Rightarrow E=1^2+2^2+3^2+4^2+...+99^2\)
\(\Rightarrow E=\dfrac{99.\left(99+1\right)\left(2.99+1\right)}{6}\)
\(\Rightarrow E=\dfrac{99.100.199}{6}\)
\(\Rightarrow E=33.50.199=328350\)
E = 1 x 1 + 2 x 2 + 3 x 3 + 4 x 4 +...+ 99 x 99
E = 1x(2-1) + 2 x (3-1)+...+ 99 x (100 -1)
D = 1 x 2 - 1 + 2 x 3 - 2 +...+ 99 x 100 - 99
D = 1x2 + 2 x 3 +...+ 99 x 100 - ( 1 + 2 +...+ 99)
Đặt A = 1x2 + 2 x 3 +...+ 99 x 100
B = 1 + 2 + ...+ 99
1x2 x 3 = 1x2x3
2x3x3 = 2x 3 x (4-1) = 2x3x4 - 1x2x3
3 x 4 x 3 = 3 x 4 x ( 5 - 2) = 3 x 4 x 5 - 2 x 3 x 4
................................................
99 x 100 x 3 = 99 x 100 x (101 - 98) = 99x100x101 - 98 x 99 x 100
Cộng vế với vế ta có: 3A = 99 x 100 x 101
A = 99 x 100 x 101 : 3 = 333300
B = 1 + 2 + 3 + ...+ 99
B = (99 + 1).[(99 -1):1 +1]:2 = 4950
E = 33300 - 4950 = 328350
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\(100-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{100}\right)\)
\(=(1-1)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+...\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}+\frac{2}{3}...+\frac{99}{100}\)
Đặt \(A=\frac{1}{1.100}+\frac{1}{2.99}+...+\frac{1}{99.2}+\frac{1}{100.1}\)
\(\Rightarrow\frac{A}{2}=\frac{1}{1.100}+\frac{1}{2.99}+...+\frac{1}{49.52}+\frac{1}{50.51}\)
\(\frac{101A}{2}=\frac{101}{1.100}+\frac{101}{2.99}+...+\frac{101}{49.52}+\frac{101}{50.51}\)
\(\frac{101A}{2}=\frac{1+100}{1.100}+\frac{2+99}{2.99}+...+\frac{49+52}{49.52}+\frac{50+51}{50.51}\)
\(\frac{101A}{2}=\frac{1}{100}+1+\frac{1}{99}+\frac{1}{2}+...+\frac{1}{52}+\frac{1}{49}+\frac{1}{51}+\frac{1}{50}\)
Do đó: \(\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}{A}=\frac{101}{2}\).