kgdc.net
当前位置:首页 >> 1x2+2x3+3x4 >>

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

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;}

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*2+2*3+3*4=2+6+12=20

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 1x2+2x3+3x4+…+10x11 =10x(10+1)x[(10x2+1)+3]/6 =110x4 =440

一般的,有: (n-1)n(n+1) =n^3-n {n^3}求和公式:Sn=[n(n+1)/2]^2 {n}求和公式:Sn=n(n+1)/2 1x2x3+2x3x4+3x4x5+....+7x8x9 =2^3-2+3^3-3+...+8^3-8 =(2^3+3^3+...+8^3)-(2+3+...+8) =[(8*9/2)^2-1]-8*9/2+1 =1260

#include int main() { int i,sum=0; for(i=1;i

1x2+2x3+3x4+…+n(n+1)  =1^2+1+2^2+2+3^2+3+…+n^2+n  =(1^2+2^2+3^2+…+n^2)+(1+2+3+…+n)  =1/6*n(n+1)(2n+1)+1/2*n(n+1)  =1/6*n(n+1)(2n+1+3)(提取公因式)  =1/3*n(n+1)(n+2)

1x2+2x3+3x4+4x5+...+n(n+1) =(1^2+2^2+……n^2)+(1+2+3+……n) =n(n+1)(2n+1)/6+(1+n)xn/2 =n(n+1)(n+2)/3

1×2+2×3+3×4+……+59×60 =1²+1+2²+2+3²+3+……+59²+59 =1/6×59×(59+1)×(2×59+1)+(1+59)×59÷2 =1/6×59×60×119+60×59÷2 =59×1190+59×30 =59×1220 =71980 希望帮到你 望采纳 谢谢 加油

网站首页 | 网站地图
All rights reserved Powered by www.kgdc.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com