# Program to print the sum of the given nth term

Given the value of the n. You have to find the sum of the series where the nth term of the sequence is given by:

Tn = n2 – ( n – 1 )2

Examples :

```Input : 3
Output : 9

Explanation: So here the tern of the sequence upto n = 3 are:
1, 3, 5 And hence the required sum is = 1 + 3 + 5 = 9

Input : 6
Output : 36
```

Simple Approach
Just use a loop and calculate the sum of each term and print the sum.

## C++

 `// CPP program to find summation of series ` `#include ` `using` `namespace` `std; ` ` `  `int` `summingSeries(``long` `n) ` `{ ` `    ``// use of loop to calculate ` `    ``// sum of each term ` `    ``int` `S = 0;  ` `    ``for` `(``int` `i = 1; i <= n; i++)  ` `        ``S += i * i - (i - 1) * (i - 1); ` `     `  `    ``return` `S; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `n = 100; ` `    ``cout << ``"The sum of n term is: "` `        ``<< summingSeries(n) << endl; ` `    ``return` `0; ` `} `

## Java

 `// JAVA program to find summation of series ` `import` `java.io.*; ` `import` `java.math.*; ` `import` `java.text.*; ` `import` `java.util.*; ` `import` `java.util.regex.*; ` ` `  `class` `GFG  ` `{ ` ` `  `    ``// function to calulate sum of series ` `    ``static` `int` `summingSeries(``long` `n) ` `    ``{ ` `        ``// use of loop to calculate ` `        ``// sum of each term ` `        ``int` `S = ``0``;  ` `        ``for` `(i = ``1``; i <= n; i++)  ` `            ``S += i * i - (i - ``1``) * (i - ``1``);      ` `         `  `        ``return` `S; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `n = ``100``; ` `        ``System.out.println(``"The sum of n term is: "` `+  ` `                            ``summingSeries(n)); ` `    ``} ` `} `

## Python3

 `# Python3 program to find summation ` `# of series ` ` `  `def` `summingSeries(n): ` ` `  `    ``# use of loop to calculate ` `    ``# sum of each term ` `    ``S ``=` `0` `    ``for` `i ``in` `range``(``1``, n``+``1``):  ` `        ``S ``+``=` `i ``*` `i ``-` `(i ``-` `1``) ``*` `(i ``-` `1``) ` `     `  `    ``return` `S ` ` `  `# Driver Code ` `n ``=` `100` `print``(``"The sum of n term is: "``,  ` `           ``summingSeries(n), sep ``=` `"") ` `# This code is contributed by Smitha. `

## C#

 `  `  `// C# program to illustrate... ` `// Summation of series ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `    ``// function to calculate sum of series ` `    ``static` `int` `summingSeries(``long` `n) ` `    ``{ ` ` `  `        ``// Using the pow function calculate ` `        ``// the sum of the series ` `        ``return` `(``int``)Math.Pow(n, 2); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String[] args) ` `    ``{ ` `        ``int` `n = 100; ` `        ``Console.Write(``"The sum of n term is: "` `+  ` `                        ``summingSeries(n)); ` `    ``} ` `} ` ` `  `// This code contribute by Parashar... `

## PHP

 ` `

Output:

```The sum of n term is: 10000
```

Time complexity – O(N)
Space complexity – O(1)

Efficient Approach
Use of mathematical approach can solve this problem in more efficient way.

Tn = n2 – (n-1)2

Sum of the series is given by (S) = SUM( Tn )

LET US TAKE A EXAMPLE IF
N = 4
It means there should be 4 terms in the series so

1st term = 12 – ( 1 – 1 )2
2nd term = 22 – ( 2 – 1 )2
3th term = 32 – ( 3 – 1 )2
4th term = 42 – ( 3 – 1 )2

SO SUM IS GIVEN BY = (1 – 0) + (4 – 1) + (9 – 4) + (16 – 9)
= 16

FROM THIS WE HAVE NOTICE THAT 1, 4, 9 GET CANCELLED FROM THE SERIES
ONLY 16 IS LEFT WHICH IS EQUAL TO THE SQUARE OF N

So from the above series we notice that each term gets canceled from the next term, only the last term is left which is equal to N2.

## C++

 `// CPP program to illustrate... ` `// Summation of series ` ` `  `#include ` `using` `namespace` `std; ` ` `  `int` `summingSeries(``long` `n) ` `{ ` `    ``// Sum of n terms is n^2 ` `    ``return` `pow``(n, 2); ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `n = 100; ` `    ``cout << ``"The sum of n term is: "` `         ``<< summingSeries(n) << endl; ` `    ``return` `0; ` `} `

## Java

 `// JAVA program to illustrate... ` `// Summation of series ` ` `  `import` `java.io.*; ` `import` `java.math.*; ` `import` `java.text.*; ` `import` `java.util.*; ` `import` `java.util.regex.*; ` ` `  `class` `GFG  ` `{ ` ` `  `    ``// function to calculate sum of series ` `    ``static` `int` `summingSeries(``long` `n) ` `    ``{ ` ` `  `        ``// Using the pow function calculate ` `        ``// the sum of the series ` `        ``return` `(``int``)Math.pow(n, ``2``); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `n = ``100``; ` `        ``System.out.println(``"The sum of n term is: "` `+  ` `                            ``summingSeries(n)); ` `    ``} ` `} `

## Python3

 `# Python3 program to illustrate... ` `# Summation of series ` `import` `math ` ` `  `def` `summingSeries(n): ` ` `  `    ``# Sum of n terms is  n^2 ` `    ``return` `math.``pow``(n, ``2``) ` ` `  `# Driver Code ` `n ``=` `100` `print` `(``"The sum of n term is: "``,  ` `        ``summingSeries(n)) ` `# This code is contributed by mits. `

## C#

 `// C# program to illustrate... ` `// Summation of series ` `using` `System; ` ` `  `class` `GFG  ` `{ ` `    ``// function to calculate sum of series ` `    ``static` `int` `summingSeries(``long` `n) ` `    ``{ ` `        ``// Using the pow function calculate ` `        ``// the sum of the series ` `        ``return` `(``int``)Math.Pow(n, 2); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 100; ` `           ``Console.Write(``"The sum of n term is: "` `+  ` `                              ``summingSeries(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by nitin mittal. `

## PHP

 ` `

Output:

`The sum of n term is: 10000`

Time complexity – O(1)
Space complexity – O(1)

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