题解 | #等差数列#
等差数列
http://www.nowcoder.com/practice/f792cb014ed0474fb8f53389e7d9c07f
//HJ100 等差数列
#include <stdio.h>
int main(void) {
int n = 0;
while (~scanf("%d", &n)) {
//等差求和：sn=n*a1+((n*(n-1)/2)*d
printf("%d\n", n * 2 + ((n * (n - 1) / 2) * 3));
}
return 0;
}
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