题解 | #等差数列#
等差数列
http://www.nowcoder.com/practice/f792cb014ed0474fb8f53389e7d9c07f
等差数列前n项求和
n-->项数
a1-首项
an->末项
sum->和
#include<stdio.h>
int main()
{
int n = 0;
int a1 = 2;
int an = 0;
int sum = 0;
while(scanf("%d",&n) != EOF)
{
an = a1 + (n-1)*3;
sum = n * (a1 + an) / 2;
printf("%d\n",sum);
}
}
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