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1x2+2x3+3x4

=1*(1+1)+2*(2+1)+…+99*(99+1) =1²+…+99²+1+…+99 =99*(99+1)(2*99+1)/6+99*(99+1)/2 =33*50*199+33*50*3 =33*50*202 =333300

因为n(n+1)=n平方+n 原式=(1平方+2平方+……n平方)+(1+2+3……n) =n(n+1)(2n+1)/6+n(n+1)/2为标准答案 注:Sn(1平方+2平方+……n平方)的证明: (a+1)³-a³=3a²+3a+1(即(a+1)³=a³+3a²+3a+1) a=1时:2³...

1 对1~99进行遍历。 2 对每个值,计算该值与该值加一的乘积。 3 将乘积累加到加和变量上。 4 输出结果。 代码: #include int main(){ int i, s; for(i = s = 0; i < 100; i ++) s+=i*(i+1); printf("%d\n",s); return 0;}

1x2+2x3+3x4+…+n(n+1) =1x(1+1)+2x(2+1)+3x(3+1)+…n(n+1) =(1^2+2^2+3^2+…+n^2)+(1+2+3+…+n) =n(n+1)(2n+1)/6+n(n+1)/2 =n(n+1)[(2n+1)+3]/6

n(n+1) =(1/3) { n(n+1)(n+2) - (n-1)n(n+1) } 1x2+2x3+3x4+...99x100 = 1x2 + (1/3) { (2x3x4 - 1x2x3) + (3x4x5 - 2x3x4) +...+(99x100x101 - 98x99x100) } = 1x2 + (1/3) { 99x100x101 -1x2x3 } = (1/3) 99x100x101 =333300

1x2+2x3+3x4+…+n(n+1) =1x(1+1)+2x(2+1)+3x(3+1)+…n(n+1) =(1^2+2^2+3^2+…+n^2)+(1+2+3+…+n) =n(n+1)(2n...

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