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.
Tn = n2 - (n-1)2
Find the sum of all of the terms of series as Sn mod (109 + 7) and,
Sn = T1 + T2 + T3 + T4 + …… + Tn
Input: 229137999Output: 218194447
Explanation
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 = n2The 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
Example
的中文翻译为：
示例
#include using namespace std;#define mod 1000000007int main() { long long n = 229137999; cout << ((n%mod)*(n%mod))%mod; return 0;}
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