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S = 1x2 + 2x3 + 3x4 + ……………… + 11x12 + 12x13
3S=1x2x3 + 2x3x3 + 3x4x3+ ………. + 11x12x3 + 12x13x3
Ta lấy K = 1x2x3 +2x3x4 + 3x4x5 + …… + 11x12x13 + 12x13x14
- 3S = 1x2x3 + 2x3x3 + 3x4x3+ ……… + 11x12x3 + 12x13x3
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K – 3S = 0 + 2x3x1 + 3x4x2 + …… .. + 11x12x10 + 12x13x11
K – 3S = K – 12x13x14
Từ đó suy ra: 3S = 12x13x14
S = 4x13x14 = 728
Cách 2:
S x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + …. + 11x12x(13-10) + 12x13x(14-11)
S x 3 = 1x2x3 + 2x3x4 – 2x3x1 + 3x4x5 – 3x4x2 + …..+ 11x12x13 – 11x12x10 +12x13x14 – 12x13x11
S x 3 = 12 x 13 x14
S = 4 x 13 x 14
S = 728
ai k minh minh k lai cho
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Đặt A=1x2+2x3+3x4+...+2016x2017
=>3A=3x1x2+3x2x3+3x3x4+...+3x2016x2017
=>3A=(3-0)x1x2+(4-1)x2x3+(5-2)x3x4+...+(2018-2015)x2016x2017
=>3A=1x2x3-0x1x2+2x3x4-1x2x3+3x4x5-2x3x4+...+2016x2017x2018-2015x2016x2017
=>3A=2016x2017x2018
=>A=\(\frac{2016\times2017\times2018}{3}\)(tự tính nha)
S = 1x2 + 2x3 + 3x4 + 4x5 + ... + 2016x2017
3S = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 2016x2017x(2018-2015)
3S = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 2016x2017x2018 - 2015x2016x2017
3S = 2016x2017x2018
S = 1/3 x 2016x2017x2018.
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A = 1 x 2 + 2 x 3 + 3 x 4 + ... + 99 x 100
3A = 1 x 2 x (3 - 0) + 2 x 3 x (4 - 1) + 3 x 4 x (5 - 2) + ... + 99 x 100 x (101 - 98)
3A = 1 x 2 x 3 - 0 x 1 x 2 + 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
3A = (1 x 2 x 3 + 2 x 3 x 4 + 3 x 4 x 5 + ... + 99 x 100 x 101) - (0 x 1 x 2 + 1 x 2 x 3 + 2 x 3 x 4 + ... + 98 x 99 x 100)
3A = 99 x 100 x 101
A = 33 x 100 x 101
A = 333300
A = 1x2+2x3+3x4+4x5+....+99x100
3A=1x2x(3-0)+2x3x(4-1)+3x4x(5-2)+...+99x100x(101-98)
3A= 1x2x3-0x1x2+2x3x4-1x2x3+3x4x5-2x3x4+...+99x100x101-98x99x100
3A= 99x100x101
A=999900 : 3 = 333300
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Gọi biểu thức trên là A, ta có :
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 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.
A x 3 = 99x100x101
A = 99x100x101 : 3
A = 333300
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1 \(\times\) 2 \(\times\) 3 = 1 \(\times\) 2 \(\times\) 3
2 \(\times\) 3 \(\times\) 3 = 2 \(\times\) 3 \(\times\) ( 4 -1) = 2 \(\times\) 3 \(\times\) 4 - 1 \(\times\) 2 \(\times\) 3
3 \(\times\) 4 \(\times\) 3 = 3 \(\times\) 4 \(\times\) ( 5 -2) = 3 \(\times\) 4 \(\times\) 5 - 2 \(\times\) 3 \(\times\) 4
4 \(\times\) 5 \(\times\) 3 = 4 \(\times\) 5 \(\times\) ( 6- 3) = 4 \(\times\) 5 \(\times\) 6 - 3 \(\times\) 4 \(\times\) 5
..................................................................................
99\(\times\)100\(\times\)3 = 99\(\times\)100\(\times\)(101-98) =99\(\times\)100\(\times\)101 - 98\(\times\)99\(\times\)100
Cộng vế với vế ta được:
1\(\times\)2\(\times\)3 + 2\(\times\)3\(\times\)3 + 3\(\times\)4\(\times\)3+ ...+99\(\times\)100\(\times\)3 = 99\(\times\)100\(\times\)101
(1\(\times\)2 + 2\(\times\)3 + 3\(\times\)4 +...+99\(\times\)100)\(\times\)3 = 99\(\times\)100\(\times\)101
1\(\times\)2 + 2\(\times\)3 + 3\(\times\)4+...+99\(\times\)100 = (99 \(\times\)100 \(\times\)101):3
1\(\times\)2 + 2\(\times\)3 + 3\(\times\)4+...+99\(\times\)100 = 333 300
Ta có :
\(\dfrac{3}{1\times2}-\dfrac{5}{2\times3}+\dfrac{7}{3\times4}-\dfrac{9}{4\times5}+...-\dfrac{25}{12\times13}+\dfrac{27}{13\times14}\)
\(=\dfrac{2+1}{1\times2}-\dfrac{3+2}{2\times3}+\dfrac{4+3}{3\times4}-...+\dfrac{14+13}{13\times14}\)
\(=1+\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{4}-...+\dfrac{1}{13}+\dfrac{1}{14}\)
\(=1+\dfrac{1}{14}=\dfrac{15}{14}\)