总结
There are many types of series in mathematics which can be solved easily in C programming. This program is to find the sum of following of series in C program.
T<sub>n</sub> = n<sup>2</sup> - (n-1)<sup>2</sup>
Find the sum of all of the terms of series as Sn mod (109 + 7) and,
Sn = T1 + T2 + T3 + T4 + ...... + Tn
Input: 229137999 Output: 218194447
Tn can be expressed as 2n-1 to get it
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As we know ,
=> 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
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#include <iostream>
using namespace std;
#define mod 1000000007
int main() {
long long n = 229137999;
cout << ((n%mod)*(n%mod))%mod;
return 0;
}以上就是C/C++程序：计算以n的平方减去（n-1）的平方为第n项的序列的和的详细内容，更多请关注php中文网其它相关文章！
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