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1 ... 1/1 x 1 + 1/2 x 2 + 1/3 x 3 + ... + 1/100 x 100
1 ... 1+1/2x2+1/3x3+...+1/100x100
1=1/1x1+1/2x2+1/3x3+...+1/100x100
Chứng minh:
C=\(\dfrac{1}{2x2}\)+\(\dfrac{1}{3x3}\)+\(\dfrac{1}{4x4}\)+.....+\(\dfrac{1}{100x100}\)<1
\(C=\dfrac{1}{2\times2}+\dfrac{1}{3\times3}+\dfrac{1}{4\times4}+...+\dfrac{1}{100\times100}\\ C< \dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{99\times100}\\ C< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ C< 1-\dfrac{1}{100}=\dfrac{99}{100}< 1\)
\(A=\dfrac{1}{2\times2}+\dfrac{1}{3\times3}+...+\dfrac{1}{100\times100}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=1-\dfrac{1}{100}=\dfrac{99}{100}\)
Quy đồng 99/100 với 3/4, ta có:
\(\dfrac{99}{100}=\dfrac{396}{400};\dfrac{3}{4}=\dfrac{300}{400}\)
So sánh A với 3/4: \(\dfrac{99}{100}>\dfrac{3}{4}\left(\dfrac{396}{400}>\dfrac{300}{400}\right)\)
Tính
A=1x2x3+2x3x3+3x4x3+4x5x3+....+98x99x3
B=1x2+2x3+3x4+4x5+...+98x99
C=1x1+2x2+3x3+4x4+5x5+...+98x98
A=1.2.3+2.3(4-1)+3.4(5-2)+4.5(6-3)+....+98.99(100-97) "." la dau nhan
A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+....+98.99.100-97.98.99
A=1.2.3+98.99.100
A= 970206
Ta có : B = 1.2 + 2.3 + 3.4 + ..... + 98.99
=> 3B = 0.1.2 + 1.2.3 - 1.2.3 + ...... + 98.99.100
=> 3B = 98.99.100
=> B = \(\frac{98.99.100}{3}\) = 323400
Ta có: \(S=1+\frac{1}{2x2}+\frac{1}{3x3}+.....+\frac{1}{10x10}\)
Ta có: 1/2x2 < 1/1x2
1/3x3 < 1/2x3
1/4x4 < 1/3x4
.......................
1/10x10 < 1/9x10
=> S< 1+1/1x2+1/2x3+1/3x4+.....+1/9x10
=> S<1+(1-1/10)
=> S < 1+9/10
=> S < 19/10 < 2
Vậy S<2