# C/C++ 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:

T_{n}= n^{2}- (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,

S_{n}= T_{1}+ T_{2}+ T_{3}+ T_{4}+ ...... + T_{n}

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, T_{n}= n^{2}- (n-1)^{2}Or, T_{n}= n^{2}- (1 + n^{2}- 2n) Or, T_{n}= n^{2}- 1 - n^{2}+ 2n Or, T_{n}= 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 = n^{2}

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 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!

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