程序代码：

unsigned foo( unsigned n )
{
    unsigned sum = 0;
    for( unsigned i=0; i!=n; ++i )
        sum += (2*i+1)*(2*i+1) + (2*i+2)*(2*i+2)*(2*i+2); // 你自己加上判断sum是否溢出的逻辑
    return sum;
}

#include <stdio.h>
#include <assert.h>

int main( void )
{
    assert( foo(0) == 0 );
    assert( foo(1) == 9 );
    assert( foo(2) == 82 );
    assert( foo(214) == 4246907254 );
}

程序代码：

unsigned bar( unsigned n )
{
    // 你自己加上判断sum是否溢出的逻辑
    unsigned lhs = n*(4*n*n-1) / 3;
    unsigned rhs = n*n*(n+1)*(n+1) * 2;
    return lhs + rhs;
}

#include <stdio.h>
#include <assert.h>

int main( void )
{
    assert( bar(0) == 0 );
    assert( bar(1) == 9 );
    assert( bar(2) == 82 );
    assert( bar(214) == 4246907254 );
}