C/C++程序：计算以n的平方减去（n-1）的平方为第n项的序列的和

数学中有很多类型的级数，可以在C编程中轻松解决。该程序是在 C 程序中求级数的以下总和。

T<sub>n</sub> = n<sup>2</sup> - (n-1)<sup>2</sup>

求级数所有项的总和，即 Sn mod (10⁹ + 7) 和，

S_n = T ₁ + T₂ + T₃ + T₄ + ...... + T_n

Input: 229137999
Output: 218194447

说明

Tn可以表示为2n-1来得到

正如我们知道，

=> 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

的中文翻译为：

示例

#include <iostream>
using namespace std;
#define mod 1000000007
int main() {
   long long n = 229137999;
   cout << ((n%mod)*(n%mod))%mod;
   return 0;
}

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