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

• Last Updated : 05 Dec, 2018

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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

 `// CPP program to find sum of given``// series.``#include ``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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