题解 | #等差数列#
等差数列
http://www.nowcoder.com/practice/f792cb014ed0474fb8f53389e7d9c07f
C题解：
| n = 1 | n = 2 | n = 3 | n = 4 | n = 5 |
|---|---|---|---|---|
| 2 | 5 | 8 | 11 | 14 |
规律：
value = 3 * n - 1
2到3n-1的求和：
(2 + (3n - 1)) * n / 2 = n * (3n + 1) / 2
因此题解如下：
int main() {
int n = 0;
while(scanf("%d", &n) != -1) {
int sum = n * (3 * n + 1) / 2;
printf("%d\n", sum);
}
return 0;
}
