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# 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 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!

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