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32.求1+(1+2)+(1+2+3)+(1+2+3+4)+……的前n项的和

#include<iostream>using namespace std;int QiuHe(int);int main(){    int n;    int sum=0;    cout<<"please input an number : "<<endl;    cin>>n;    for(int i=1;i<=n;i++)    {        sum+=QiuHe(i);    }    cout<<"the mount of the series is : "<<sum<<endl;    return 0;}int QiuHe(int index){    int sum1=0;    for(int j=1;j<=index;j++)    {        sum1+=j;    }    return sum1;}