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

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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 Sn mod (109 + 7), where Sn 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. Tn 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 ?Tn.

?Tn = ?(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,

?Tn = (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!


Last Updated : 19 Jul, 2022
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