题解 | #等差数列#
等差数列
http://www.nowcoder.com/practice/f792cb014ed0474fb8f53389e7d9c07f
#include<bits/stdc++.h>
using namespace std;
int main() {
int n;
while(cin>>n) {
int sum = n*2 + n*(n-1)/2*3; //等差数列求和公式Sn = n·a1 + n(n-1)/2 * d，忘记的同学也可以自行百度回忆一下
cout<<sum<<endl; //输出等差数列前N项的和
}
return 0;
}
