大家帮忙看哈啊。。。我想了好久，没辙了，求指点
第一项k1=0，k2=k1+2^n(n=1,2,3...),
sum=k1^2+k2^2+k1*k2;
帮忙写一下代码。。。我想了好久，没辙了，求指点

[解决办法]

C/C++ code

//k1=0，k2=k1+2^n(n=1,2,3...,15),//sum=k1^2+k2^2+k1*k2;#include <stdio.h>int k1,k2;int n,sum;int main() {    for (n=1;n<=15;n++) {        k1=0;        k2=k1+(1<<n);        sum=k1*k1+k2*k2+k1*k2;        printf("n==%2d,sum==%d\n",n,sum);    }    return 0;}//n== 1,sum==4//n== 2,sum==16//n== 3,sum==64//n== 4,sum==256//n== 5,sum==1024//n== 6,sum==4096//n== 7,sum==16384//n== 8,sum==65536//n== 9,sum==262144//n==10,sum==1048576//n==11,sum==4194304//n==12,sum==16777216//n==13,sum==67108864//n==14,sum==268435456//n==15,sum==1073741824