tính nhanh
Z=1.2+2.3+3.4+...........+99.100
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Ta có 3 x S = 1 x 2 x 3 + 2 x 3 x 3 + 3 x 4 x 3 + ... + 99 x 100 x 3
3 x S = 1 x 2 x (3 - 0) + 2 x 3 x (4 - 1) + 3 x 4 x (5 - 2) + ... + 99 x 100 x (101 - 98)
3 x S = 1 x 2 x 3 + 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 100
=> 3 x S = 99 x 100 x 101
=> A = 33 x 100 x 101 = 333300
S = 1.2 + 2.3 + 3.4 + 4.5 + ..... + 99.100
3S=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+.....+99.100.(101-98)
3S = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ..... + 99.100.101
3S=99.100.101
S=99.100.101/3
S=333300
1. ta có :
\(3^2+4^2=5^{x-1}\)
\(25=5^{x-1}\)
\(5^2=5^{x-1}\)
=> x = 3
Ta có : S = 1.2 + 2.3 + 3.4 + ..... + 99.100
=> 3S = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ..... + 99.100.101
=> 3S = 99.100.101
=> S = 99.100.101/3
=> S = 333300
A=3(1/1.2+1/2.3+...+1/99.100)
A=3(1-1/2+1/2-1/3+...+1/99-1/100)
A=3(1-1/100)
A=3 . 99/100
A= 297 /100
5B= 1.2.3.4.5+2.3.4.5.5+....+97.98.99.100.5
=1.2.3.4.5+2.3.4.5.6 -1.2.3.4.5+...+-96.97.98.99
=97.98.99.100.101=9505049400
=> B=1901009880
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)
\(A=\frac{1}{9}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(A=\frac{1}{9}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=\frac{1}{9}.\left(1-\frac{1}{100}\right)\)
\(A=\frac{1}{9}.\frac{99}{100}\)
\(A=\frac{11}{100}\)
A = 9/1.2 + 9/2.3 + 9/3.4 +...+ 9/98.99 + 9/99.100
= 9. (1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/98 - 1/99 + 1/99 - 1/100)
= 9. (1 - 1/100)
= 9 . 99/100
= 891/100
3S=1*2*3+2*3*(4-1)+...+99*100*(101-98)
=1*2*3+2*3*4-1*2*3+...+99*100*101-99*100*98
=99*100*101
=>S=33*100*101=333300
1.2+2.3+3.4+4.5+...+99.100
=1.2.3+2.3.3+3.4.3+4.5.3+...+99.100.3
=1.2.3+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+...+99.100.(101-98)
=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+99.100.101-98.99.100
=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+4.5.6-4.5.6+...+99.100.101
=99.100.101=999900
=999900:3=333300
3Z = 1.2.3 + 2.3.3 + 3.3.4 + .... + 3.99.100
Z= 1.2.3 + 2.3.4 + 3.4.5 + .... + 99.100.101
=> Z - 3Z = 1.2.3 - 1.2.3 + 2.3.(4-3) + 3.4 ( 5-3) + .... + 99.100 ( 101 -3)
= 1.2.3 + 2.3.4 + .... + 98.99.100
=> Z -3Z = Z - 99.100.101
=> Z = 99.100.101/3 = 333300
\(Z=1.2+2.3+3.4+...+99.100\)
\(3Z=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)\)
\(3Z=1.2.3+2.3.4-1.2.3+3.4.5-2.4.5+...+99.100.101-98.99.100\)
\(3Z=99.100.101\)
\(Z=\frac{99.100.101}{3}=\frac{999900}{3}=333300\)