1222222 X2212 + [99X100]
6.Cho a, b, c và x, y, z là các số khác nhau và khác không chứng minh rằngNếu: 0=++zcybxavà 1=++czbyaxthì 1222222=++czbyax
Tính tổng:
A = 1x2+3x4+4x5+...+99x100
B = 1x22+2x32+3x42+4x52+...+99x1002
Tinh 1x2+2x3+3x4+...+99x100
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
eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee2
Tinh 1x2+2x3+3x4+...+99x100
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
1x2 + 2x3 + 3x4 + 4x5 + .... +99x100
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
E=1x2+2x3+...+99x100
E=1x2+2x3+...+99x100
E x 3 = 1 x 2 x 3 + 2 x 3 x 3 + ...... + 99 x 100 x 3
E x 3 = 1 x 2 x ( 3 - 0 ) + 2 x 3 x ( 4 - 1 ) + ....... + 99 x 100 x ( 101 - 98 )
E x 3 = 1 x 2 x 3 - 1 x 2 x 0 + 2 x 3 x 4 - 2 x 3 x 1 + ....... + 99 x 100 x 101 - 99 x 100 x 98
E x 3 = 99 x 100 x 101
E = 99 x 100 x 101 : 3
E = 333330
Nếu mình đúng các bạn k mình nhé
3E=1x2x3+2x3x3+...+99x100x3
3E=1x2x3+2x3x(4-1)+...+99x100x(101-98)
3E=1x2x3+2x3x4-1x2x3+...+99x100x101-98x99x100
3E=99x100x101
E=99x100x101:3
E=333300
1x2+1x2x3+1x2x3x4+1x2x3x4x5+....+1x2x3...99x100
[1x1]x....x[99x100]=?
1x2+3x4+5x6+.............................+99x100