tính tổng \(S=1\times2+2\times3+3\times4+4\times5....+99\times100\)
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\(=\frac{1.2}{99.100}\)
\(=\frac{2}{9900}=\frac{1}{4950}\)
\(1,\\ =\dfrac{2-1}{1\times2}+\dfrac{3-2}{2\times3}+\dfrac{4-3}{3\times4}+\dfrac{5-4}{4\times5}+.....+\dfrac{99-98}{98\times99}+\dfrac{100-99}{99\times100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{100-1}{100}=\dfrac{99}{100}\)
\(2,=\dfrac{13-11}{11\times13}+\dfrac{15-13}{13\times15}+....+\dfrac{21-19}{19\times21}+\dfrac{23-21}{21\times23}\\ =\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+....+\dfrac{1}{19}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{23}\\ =\dfrac{1}{11}-\dfrac{1}{23}\\ =\dfrac{23-11}{11\times23}=\dfrac{12}{253}\)
@seven
a: 1/1*2+1/2*3+...+1/99*100
=1-1/2+1/2-1/3+...+1/99-1/100
=1-1/100
=99/100
b: 2/11*13+2/13*15+...+2/21*23
=1/11-1/13+1/13-1/15+...+1/21-1/23
=1/11-1/23
=12/253
C=1x2+2x3+3x4 +......+99x100
<=>3xC=1x2x3+2x3x3+.....+99x100x3
<=>3xC=1x2x3+2x3x(4-1)+.....+99x100x(101-98)
<=>3xC=1x2x3-1x2x3+2x3x4-2x3x4+......-98x99x100+99x100x101
<=>3xC=99x100x101 <=> C=99x100x101 :3 <=> C= 333300
\(C=1.2+2.3+3.4+...+99.100\)
=> 3C = 1.2.3 + 2.3.3 + 3.4.3 + ...+ 99.100.3
3C = 1.2.3 + 2.3.(4-1) + 3.4.(5-2) + ...+ 99.100.(101-98)
3C = 1.2.3 + 2.3.4 -1.2.3 + 3.4.5 - 2.3.4 + ...+ 99.100.101-98.99.100
\(3C=\left(1.2.3+2.3.4+3.4.5+...+99.100.101\right)-\left(1.2.3+2.3.4+...+98.99.100\right)\)
\(3C=99.100.101\Rightarrow C=\frac{99.100.101}{3}=333300\)
Dấu . là dấu nhân
3C = 3 x C
P = 2 x 3 + 3 x 4 + ...+ 99 x 100
=> 3 x P = 2 x 3 x 3 + 3 x 4 x 3 + ....+ 99 x 100 x 3
3 x P = 2 x 3 x ( 4-1) + 3 x 4 x (5-2) + ...+ 99 x 100 x ( 101 -98)
3 x P = 2 x 3 x 4 - 1 x 2 x 3 + 3 x 4 x 5- 2 x 3 x 4 + ...+ 99 x 100 x 101 - 98 x 99 x 101
3 x P = ( 2 x 3 x 4 + 3 x 4 x 5 + ...+ 99 x 100 x 101) - ( 1 x 2 x 3 + 2 x 3 x 4 + ...+ 98 x 99 x 101)
3 x P = 99 x 100 x 101 - 1 x 2 x 3
\(P=\frac{99x100x101-1x2x3}{3}=333298\)
p=2.3+3.4+4.5+5.6+...+99.100
3p=2.3.3+3.4.3+4.5.3+5.6.3+...+99.100.3
3p=2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+5.6.(7-4)+...+99.100.(101-98)
3p=2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+5.6.7-4.5.6+99.100.101-98.99.100
3p=98.99.100-1.2.3
p=\(\frac{98.99.100-1.2.3}{3}=323398\)
dau . la dau x
a/ 1.3.2.4.3.5.4.6.5.7/2.2.3.3.4.4.5.5.6.6=1.7/2.6=7/12
b/ ab.aba=abab
aba=abab:ab
aba=101
=>a=1 b=0
aabb : ab = 99 hay ab x 99 = aabb hay ab x100 – ab = aabb
Ta có phép tính
__ ab00
___ab___
aabb
b=0 hoặc b=5
Nếu b=0 thì a000 – a0 = aa00 (sai)
Nếu b=5 thì
__ a500
__a5___
aa55
a=4
c) thay a=7/6 b=6/5 thi 3 x a + 4 : b - 5/12=3.7/6+4.6/5-5/12=7/2+24/5-5/12=210/60+288/60-25/60=473/60
**** nha
\(\frac{1.3.2.4.3.5.4.6.5.7}{2.2.3.3.4.4.5.5.6.6}=\frac{\left(2.3.4.5.6\right).\left(3.4.5.7\right)}{\left(2.3.4.5.6\right).\left(2.3.4.5.6\right)}=\frac{7}{12}\)
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+....+\frac{1}{98\times99}+\frac{1}{99\times100}\)
\(=\frac{2-1}{1\times2}+\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+....+\frac{99-98}{98\times99}+\frac{100-99}{99\times100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
ta có :
S = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
S x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
S x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
S x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.
S x 3 = 99x100x101
S = 99x100x101 : 3
S = 333300