C / C ++程序，用于查找第n个项为n ^ 2 –(n-1)^ 2的级数之和
数学中有许多类型的级数，可以在C编程中轻松解决。该程序是在C程序中找到系列后继的和。
Tn = n2 - (n-1)2
找到所有序列项的总和，即Snmod（109+7），然后，
Sn=T1+T2+T3+T4+......+Tn
Input: 229137999 Output: 218194447
说明
可以将Tn表示为2n-1
据我们所知，
=> Tn = n2 - (n-1)2 =>Tn = n2 - (1 + n2 - 2n) =>Tn = n2 - 1 - n2 + 2n =>Tn = 2n - 1. find ∑Tn. ∑Tn = ∑(2n – 1) Reduce the above equation to, =>∑(2n – 1) = 2*∑n – ∑1 =>∑(2n – 1) = 2*∑n – n. here, ∑n is the sum of first n natural numbers. As known the sum of n natural number ∑n = n(n+1)/2. Now the equation is, ∑Tn = (2*(n)*(n+1)/2)-n = n2 The value of n2 can be large. Instead of using n2 and take the mod of the result. So, using the property of modular multiplication for calculating n2: (a*b)%k = ((a%k)*(b%k))%k
示例
#include <iostream>
using namespace std;
#define mod 1000000007
int main() {
long long n = 229137999;
cout << ((n%mod)*(n%mod))%mod;
return 0;
}