Open In App

C/C++ Program to Find sum of Series with n-th term as n^2 - (n-1)^2

Last Updated : 23 Jul, 2025
Comments
Improve
Suggest changes
Like Article
Like
Report

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 (10

9

+ 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

.

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 n

2

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

CPP
// CPP program to find sum of given
// series.
#include <bits/stdc++.h>
using namespace std;

#define mod 1000000007

// Function to find sum of series
// with nth term as n^2 - (n-1)^2
long long findSum(long long n)
{
    return ((n % mod) * (n % mod)) % mod;
}

// Driver code
int main()
{
    long long n = 229137999;
    cout << findSum(n);
    return 0;
}

Output
218194447

Please refer complete article on

Find sum of Series with n-th term as n^2 - (n-1)^2

for more details!


Article Tags :

Similar Reads