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a) Số số hạng: \(\frac{\left(99-1\right)}{1}+1=99\)
Tổng: \(\frac{99+1}{2}\cdot99=4950\)
b) Số số hạng: \(\frac{\left(100-2\right)}{2}+1=50\)
Tổng: \(\frac{100+2}{2}\cdot50=2550\)
c) \(S=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\)
\(3\cdot S=1\cdot2\left(3-0\right)+2\cdot3\left(4-1\right)+3\cdot4\left(5-2\right)+...+99\cdot100\left(101-98\right)\)
\(3\cdot S=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+3\cdot4\cdot5-2\cdot3\cdot4+...+99\cdot100\cdot101-98\cdot99\cdot100\)
\(3\cdot S=99\cdot100\cdot101\)
Vậy, \(S=\frac{1}{3}\cdot99\cdot100\cdot101=333300\)
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cau 1 :
Tu dau bai ta co:
X x 50 + ( 100 + 2 ) x 50 : 2 = 2700
X x 50 + 2550 = 2700
X x 50 = 2700 - 2550
X x 50 = 150
X = 150 : 50
X = 3
Bai 2
1 ) ( 99 + 0 ) x 100 : 2 = 4950
2 ) ( 99 + 1 ) x 99 : 2 = 4950
3 ) day tren co so cac so hang la :
( 99 - 1 ) : 2 + 1 = 50 ( so )
Vay tong la : ( 99 + 1 ) x 50 : 2 = 2500
lik e cho minh nha moi nguoi
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G = 1.2 + 1 + 2.3 + 1 + 3.4 + 1 + ... + 99.100 + 1
= 1.2 + 2.3 + 3.4 + ... + 99.100 + 99
⇒ 3G = 1.2.3 + 2.3.3 + 3.4.3 + ... + 99.100.3 + 99.3
= 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 99.100.(101 - 98)
= 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + 3.4.5 - ... - 98.99.100 + 99.100.101 + 297
= 99.100.101 + 297
= 1000197
⇒ G = 1000197 : 3
= 333399
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\(A=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{99}{100}=\dfrac{1}{100}\)
Dat S la tong phai tim. Ta có:
Sx3=1x2x3+2x3x3+3x4x3+....+99x100x3
=1x2x(3-0)+2x3x(4-1)+3x4x(5-2)+....+99x100x(101-98)
=(1x2x3-0x1x2)+(2x3x4-1x2x3)+(3x4x5-2x3x4)+........+99x100x101-98x99x100
=99x100x101-0x1x2=99x100x101=999900