为了找到该系列的总和，我们将首先分析该系列。
该系列是：2,6,12,20,30…
For n = 6 Sum = 112 On analysis, (1+1),(2+4),(3+9),(4+16)... (1+12), (2+22), (3+32), (4+42), can be divided into two series i.e. s1:1,2,3,4,5… andS2: 12,2,32,....
用数学公式求出第一和第二的总和
Sum1 = 1+2+3+4… , sum1 = n*(n+1)/2 Sum2 = 12+22+32+42… , sum1 = n*(n+1)*(2*n +1)/6
#include <stdio.h>
int main() {
int n = 3;
int sum = ((n*(n+1))/2)+((n*(n+1)*(2*n+1))/6);
printf("the sum series till %d is %d", n,sum);
return 0;
}输出结果
The sum of series till 3 is 20