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,
Sn = T 1 + T2 + T3 + T4 + ...... + Tn
Input: 229137999 Output: 218194447
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
#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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