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Summing the N series

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  • orders6
     + 0 comments

    I'm new to this website and this confused me as to it being so simple if you know series and realize that your essentially ding a summation from k=1 to n of n^2-(n-1)^2 which is essentially equal to n^2. In other words the function is simply n^2. So to return the value with the modulus one just has to return this into just one line: ((n%1000000007)*(n%1000000007))%1000000007

  • imogenhudson938
     + 0 comments

    That’s a great breakdown Performance Optimization Consulting USA often focuses on simplifying complex processes just like this elegant reduction of the series to a single squared term.

  • imogenhudson938
     + 0 comments

    It looks like understanding sequences step by step is crucial here, much like how structured b2b sales training London programs guide professionals through systematic learning for better results.

  • h2001203004
     + 0 comments

    Here's my solution:

    #include <bits/stdc++.h>
using namespace std;

string ltrim(const string &);
string rtrim(const string &);

/*
 * Complete the 'summingSeries' function below.
 *
 * The function is expected to return an INTEGER.
 * The function accepts LONG_INTEGER n as parameter.
 */


/*
T_n=n^2-(n-1)^2

a^2+b^2=(a+b)(a-b)

T_n = (n+n-1)(n-(n-1))=(2n-1)(1)=2n-1

S_n = T_1+T_2+...+T_n
S_n = sum_{k=1}^{n} (2k-1)
S_n = 2sum_{k=1}^{n} k - sum_{k=1}^{n} 1
S_n = 2* (n*n-1)/2 -n = n(n+1)-n = n^2
==> S_n = n^2
S_n mod (10^9+7) = n^2 mod 1000000007
*/

int summingSeries(long n) {
    // return (1LL*n*n)%1000000007; // Fail 2 testcases
    const int MOD = 1000000007;
    n%=MOD;
    return (n*n)%MOD;
}

int main()
{
    ofstream fout(getenv("OUTPUT_PATH"));

    string t_temp;
    getline(cin, t_temp);

    int t = stoi(ltrim(rtrim(t_temp)));

    for (int t_itr = 0; t_itr < t; t_itr++) {
        string n_temp;
        getline(cin, n_temp);

        long n = stol(ltrim(rtrim(n_temp)));

        int result = summingSeries(n);

        fout << result << "\n";
    }

    fout.close();

    return 0;
}

string ltrim(const string &str) {
    string s(str);

    s.erase(
        s.begin(),
        find_if(s.begin(), s.end(), not1(ptr_fun<int, int>(isspace)))
    );

    return s;
}

string rtrim(const string &str) {
    string s(str);

    s.erase(
        find_if(s.rbegin(), s.rend(), not1(ptr_fun<int, int>(isspace))).base(),
        s.end()
    );

    return s;
}

  • diyadavidpnc
     + 0 comments

    Java 8

    return (int)(((n%1000000007)*(n%1000000007))%1000000007);