x 1 x 2-x 3 x 4=x 5 x 6-x 7 x 8

n(n+1)(n+2)(n+3)/4

按题设的规律,(6)=5*6*7,(7)=6*7*8,由等式知·=(7)/(6)+1=(6*7*8)/(5*6*7)+1=13/5

解:令x1=y1+y2,x2=y1-y2,x3=y3,x4=y4f=y1^2-y2^2+y1y3-y2y3+y3y4=(y1+y3/2)^2-(y2+y3/2)^2+y3^2y3y4=z1^2-z2^

2和3因为三个连续自然数至少有一个是偶数,且大于或等于2,所以它们的乘积一定是2的倍数三个连续自然数有一个是3的倍数,所以它们的乘积一定是3的倍数

nx(n+1)=1/3[n(n+1)(n+2)-(n-1)n(n+1)]1x2+2x3+3x4+...+nx(n+1)=1/3[1x2x3-0x1x2+2x3x4-1x2x3+3x4x5-2x3x4+

读完以上材料,请你计算下列各题：（1）1×2+2×3+3×4+…+10×11（写出过程）；1×2+2×3+3×4+…+10×11=1/3（1×2×3-0×1×2）+1/3（2×3×4-1×2×3）+1

3*(1x2+2x3+3x4+...+99x100)=3*1/3*(1x2x3-0x1x2+2x3x4-1x2x3+3x4x5-2x3x4+99x100x101-98x99x100)=99x100x1

1X2+2X3+3X4+、、、、、、+nX(n+1）=(1/3)(1*2*3-0*1*2)+(1/3)(2*3*4-1*2*3)+(1/3)(3*4*5-2*3*4)+.+(1/3)[n*(n+1)(

原式=1/4（-0*1*2*3+1*2*3*4）+1/4（-1*2*3*4+2*3*4*5）+……+1/4[-（n-1)n(n+1)(n+2)+n(n+1)(n+2)(n+3)]=1/4[-0*1*2

1/2x3X4=（1/2X3-1/3X4）X1/21/3X4x5=（1/3X4-1/4X5）X1/2.1/8X9X10=（1/8X9-1/9X10）X1/2上述各式相加得：1/2x3X4十1/3X4x

1/1x2x3+1/2x3x4+1/3x4x5+1/4x5x6+...+1/8x9x10=1-1/2-1/2(1-1/3)+1/2-1/3-1/2(1/2-1/4)+1/3-1/4-1/2(1/3-1

裂项求和1/((n-1)n(n+1))=(1/2)*(1/((n-1)n)-1/(n(n+1)))接着便可算了结果我算的是(1/2)*(1/6-1/(49*50))

根据韦达定理x1x2=c>0x3x4=b>0x1+x2=-bx3+x4=-c因为两个方程都有两个正整数根x1,x2,x3,x4都是正整数因此c和b也是正整数c-b=x1x2-x1-x2=(x1-1)(

（4）+（6）=3*4*5+5*6*7=60+210=270

6/7

=1/2×(1/1×2-1/2×3+1/2×3-1/3×4+1/3×4-1/4×5+1/4×5-1/5×6)=1/2×(1/1×2-1/5×6)=7/30

（1）1x2+2x3+…+99x100+100x101==1/3x100x101x102=343400（2）1x2+2x3+3x4+…+n(n+1)(n为正整数)=1/3n(n+1)(n+2)（3）1

（1）1x2+2x3+3x4+…+10x11=1*10*11*12/3=440(2)原式=n(n+1)(n+2)/3(3)1x2x3+2x3x4+3x4x5+…+7x8x9=1x2x3×4/4-0×1

3*(1x2+2x3+3x4+...+99x100)=3*1/3*(1x2x3-0x1x2+2x3x4-1x2x3+3x4x5-2x3x4+99x100x101-98x99x100)=99x100x1

1/1x2x3+1/2x3x4+1/3x4x5+1/4x5x6+.+1/48x49x50=48*51÷(4*49*50)=306/1225