Java Program to Find sum of Series with n-th term as n^2 – (n-1)^2

We are given an integer n and n-th term in a series as expressed below:

Tn = n2 - (n-1)2

We need to find S_n mod (10⁹ + 7), where S_n is the sum of all of the terms of the given series and,

Sn = T1 + T2 + T3 + T4 + ...... + Tn

Examples:

Input : 229137999
Output : 218194447

Input : 344936985
Output : 788019571

Let us do some calculations, before writing the program. T_n can be reduced to give 2n-1 . Let’s see how:

Given, Tn = n2 - (n-1)2
Or, Tn =  n2 - (1 + n2 - 2n)
Or, Tn =  n2 - 1 - n2 + 2n
Or, Tn =  2n - 1. 

Now, we need to find ∑T_n.

∑T_n = ∑(2n – 1)

We can simplify the above formula as, ∑(2n – 1) = 2*∑n – ∑1 Or, ∑(2n – 1) = 2*∑n – n. Where, ∑n is the sum of first n natural numbers. We know the sum of n natural number = n(n+1)/2. Therefore, putting this value in the above equation we will get,

∑T_n = (2*(n)*(n+1)/2)-n = n²

Now the value of n² can be very large. So instead of direct squaring n and taking mod of the result. We will use the property of modular multiplication for calculating squares:

(a*b)%k = ((a%k)*(b%k))%k

Java

// Java program to find sum of given
// series.

public class FINDSUM {

    static long mod = 1000000007;

    public static long findSum(long n)
    {
        return ((n % mod) * (n % mod)) % mod;
    }

    public static void main(String[] args)
    {
        long n = 229137999;
        System.out.print(findSum(n));
    }
}

// Contributed by _omg

Output:

218194447

Time complexity: O(1) since performing constant operations

Auxiliary Space: O(1)

Please refer complete article on Find sum of Series with n-th term as n^2 – (n-1)^2 for more details!

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