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\(G=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\)
\(=1\left(1+1\right)+2\left(1+2\right)+3\left(1+3\right)+...+99\left(1+99\right)\)
\(=\left(1^2+2^2+3^3+...+99^2\right)+\left(1+2+3+...+99\right)\)
\(=\dfrac{99\left(99+1\right)\left(2\cdot99+1\right)}{6}+\dfrac{99\cdot100}{2}\)
=328350+4950
=333300
4/1.2 + 4/2.3 +4/3.4+...+4/99.100
=> 4/1 - 4/2 + 4/2 - 4/3 + 4/3 - 4/4+...+ 4/99 - 4/100
=>4 - 4/100
\(\frac{4}{1.2}+\frac{2}{2.3}+\frac{4}{3.4}+...+\frac{4}{99.100}\)
\(=4.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(=4.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=4.\left(1-\frac{1}{100}\right)\)
\(=4.\frac{99}{100}=\frac{99}{25}\)
3S = 1.2.3 + 2.3.3 + 1.4.3 + ... + 99.100.3
= 1.2.(4 - 1) + 3.4.(5 - 2) + ... + 99.100.(101 - 98)
= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 3.4.2 + .. + 99.100.101 - 99.100.98
=> 99.100.101 => S = 99.100.101/3 = 999900
Vậy: S = 999900
c) Đặt \(A=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\)
Ta có: \(A=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\)
\(\Leftrightarrow3A=3\cdot\left(1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\right)\)
\(\Leftrightarrow3A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+99\cdot100\cdot\left(101-98\right)\)
\(\Leftrightarrow3\cdot A=1\cdot2\cdot3-1\cdot2\cdot3+2\cdot3\cdot4-2\cdot3\cdot4+...+98\cdot99\cdot100-98\cdot99\cdot100+99\cdot100\cdot101\)
\(\Leftrightarrow3\cdot A=99\cdot100\cdot101\)
\(\Leftrightarrow A=33\cdot100\cdot101=333300\)
b) Ta có: \(1+2-3-4+...+97+98-99-100\)
\(=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(97+98-99-100\right)\)
\(=\left(-4\right)+\left(-4\right)+...+\left(-4\right)\)
\(=-4\cdot25=-100\)
a) \(1.2+2.3+3.4+...+19.20\)
\(=\dfrac{20.\left(20+1\right).\left(20+2\right)}{3}\)
\(=3080\)
b) \(9+99+999+...+999...9\left(100so9\right)\)
\(\)\(=\left(10-1\right)+\left(100-1\right)+\left(1000-1\right)+...+\left(1000...0-1\right)\left(99so0\right)\)
\(=\left(10+10^2+10^3+...10^{99}\right)+\left(-1\right).100\)
\(=\left(1+10+10^2+10^3+...10^{99}\right)+\left(-1\right).101\)
\(=\dfrac{10^{99+1}-1}{99-1}-101\)
\(=\dfrac{10^{100}-1}{98}-101\)
\(=\dfrac{10^{100}-9899}{98}\)
c) \(999.9x222...2\) (100 số 9; 100 số 2)
\(9x2=18\)
\(99x22=2178\)
\(999x222=\text{221778}\)
\(9999x2222=22217778\)
\(99999x22222=2222177778\)
\(.........\)
Theo quy luật trên ta có 100 số 9 nhân 100 số 2:
\(999.9x222...2=222...21777...78\) (99 sô 2; 1 số 1; 99 số 7; 1 số 8)
A=9/1.2+ 9/2.3+ 9/3.4+ .... +9/98.99 + 9/99/100
=9(1- 1/2 + 1/2 -1/3+...+1/99 -1/100)
=9.(1- 1/100)
=9.99/100
=891/100
A=9/1.2+9/2.3+...+9/99.100
A/9=1/1.2+1/2.3+....+1/99.100
A/9=1-1/2+1/2-1/3+....+1/99-1/100
A/9=1+(-1/2+1/2)+(-1/3+1/3)+....+(-1/99+1/99)-1/100
A/9=1-1/100
A/9=99/100
A=99/100.9=891/100
Vậy A=891/100
mik ko biết đúng hay sai mn góp ý giúp mik nha
Xét mẫu số: 1/(2x3) + 1/(3x4) + …… + 1/(99x100)
= 1/1 – 1/2 + 1/3 – 1/4 + ......... + 1/99 – 1/100
= (1 + 1/3 + ............ + 1/99) – (1/2 + 1/4 + .......... + 1/100)
= (1 + 1/3 + ............ + 1/99)+(1/2+1/4+1/6+….+1/100) – (1/2+1/4+1/6+ .......... + 1/100)x2
= (1 + 1/2 + 1/3 + 1/4 + ..... + 1/99 + 1/100) – (1 + 1/2 + 1/3 + ....... +1/50 )
= 1/51 + 1/52 + 1/53 + ............. + 1/100 (Đơn giản số trừ)
=>(1/51 + 1/52 + 1/53 + ............. + 1/100) / (1/51 + 1/52 + 1/53 + ............. + 1/100) = 1
Gọi A là biểu thức ta có:
A = 1.2+2.3+3.4+......+99.100
Gấp A lên 3 lần ta có:
A . 3 = 1.2.3 + 2.3.3 + 3.4.3 + … + 99.100.3
A . 3 = 1.2.3 + 2.3.(4 - 1) + 3.4.( 5 - 2) + … + 99.100. (101 - 98)
A . 3 = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + … + 99.100.101 - 98.99.100
A . 3 = 99.100.101
A = 99.100.101 : 3
A = 33.100.101
A = 333 300
1.2 + 2.3 + 3.4 +...+99.100 = \(\frac{99.\left(99+1\right)\left(99+2\right)}{3}=333300\)