Tính nhanh dãy số sau:
1x2+2x3+3x4+4x5+.....+199x200
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A = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3 A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98) ..................................
A x 3 = 99x100x101 A = 333300
A = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
..................................
A x 3 = 99x100x101
A = 333300
SAi rồi ! phải là 2666600 Mới đúng
Muốn biết thì bấm vào Đúng 0
Lời giải:
Gọi tổng trên là $A$
$A=2(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{199.200})$
$=2(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{200-199}{199.200})$
$=2(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{199}-\frac{1}{200})$
$=2(\frac{1}{2}-\frac{1}{200})=1-\frac{1}{100}=\frac{99}{100}$
A = \(\dfrac{2}{2.3}\) + \(\dfrac{2}{3.4}\) + \(\dfrac{2}{4.5}\) + ... + \(\dfrac{2}{199.200}\)
A = 2. (\(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + ... + \(\dfrac{1}{199.200}\))
A = 2.(\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + ... + \(\dfrac{1}{199}\) - \(\dfrac{1}{200}\))
A = 2.(\(\dfrac{1}{2}\) - \(\dfrac{1}{200}\))
A = 2. \(\dfrac{99}{200}\)
A = \(\dfrac{99}{100}\)
\(A=\frac{1.98+2.97+3.96+...+98.1}{1.2+2.3+3.4+...+98.99}=\frac{1.\left(100-2\right)+2\left(100-3\right)+3\left(100-4\right)+...+98\left(100-99\right)}{1.2+2.3+3.4+...+98.99}\)
\(A=\frac{1.100-1.2+2.100-2.3+3.100-3.4+...+98.100-98.99}{1.2+2.3+3.4+...+98.99}\)
\(A=\frac{\left(1.100+2.100+3.100+...+98.100\right)-\left(1.2+2.3+3.4+...+98.99\right)}{1.2+2.3+3.4+...+98.99}\)
\(A=\frac{100\left(1+2+3+...+98\right)}{1.2+2.3+3.4+...+98.99}-1\)
Ta có: 1+2+3+...+98=98.99:2=4851
Đặt B=1.2+2.3+3.4+...+98.99 => 3B=1.2.3+2.3.3+3.4.3+...+98.99.3 = 1.2.3+2.3.(4-1)+3.4(5-2)+...+98.99(100-97)
=> 3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+98.99.100-97.98.99 = 98.99.100
=> B=33.98.100. Thay vào A được:
\(A=\frac{100.4851}{33.98.100}-1=\frac{3}{2}-1=\frac{1}{2}\)
Gọi biểu thức trên là A, ta có :
A = 1x2 + 2x3 + 3x4 + 4x5 + ...+199x200
A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 199x100x3
A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
A x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 199x200x101 - 198x199x100.
A x 3 = 199x200x101
A = 199x200x201:3
A=2666600
Đặt A = 1 x 2 + 2 x 3 + ... + 199 x 200
3A = 1 x 2 x 3 + 2 x 3 x (4-1) + .... + 199 x 200 x (201 - 198)
3A = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 +.... + 199 x 200 x 201 - 198 x 199 x 200
3A = ( 1 x 2 x 3 - 1 x 2 x 3) + ( 2 x 3 x 4 - 2 x 3 x 4) + ....... + (198 x 199 x 200 - 198 x 199 x 200) + 199 x 200 x 201
Do đó A = 67 x 200 x 199 = 2666600
Đặt A=1x2+2x3+3x4+4x5+........+199x200
Ta có:
3A=1x2x3+2x3x3+3x4x3+.......+199x200x3
3A=1x2x3+2x3x(4-1)+3x4x(5-2)+....+199x200x(201-198)
3A=1x2x3+2x3x4-1x2x3+3x4x5-2x3x4+.............+199x200x201-198x199x200
3A=199x200x201
A=39800x201:3
A=39800x67
A=2666600
Vậy 1x2+2x3+3x4+........+199x200=2666600