# Program to print the sum of the given nth term

Last Updated : 29 Oct, 2023

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 : 3Output : 9Explanation: So here the term of the sequence upto n = 3 are:              1, 3, 5 And hence the required sum is = 1 + 3 + 5 = 9Input : 6Output : 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 calculate 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...`

## Javascript

 ``

## 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.`

## Javascript

 ``

## PHP

 ``

Output:

`The sum of n term is: 10000`

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

#### Approach#3: Using for loop

Take the value of ‘n’ from the user. Initialize the sum variable to 0. Use a for loop to iterate over the range of 1 to 2*n with a step size of 2. In each iteration, add the current value to the sum variable. After the loop, print the sum of the first n terms of the sequence.

#### Algorithm

1. Start
2. Take the value of ‘n’ from the user and store it in a variable.
3. Initialize the sum variable to 0.
4. Use a for loop to iterate over the range of 1 to 2*n with a step size of 2.
a. Add the current value to the sum variable.
5. After the loop, print the sum of the first n terms of the sequence.
End

## C++

 `#include ` `using` `namespace` `std;`   `int` `main() {` `    ``// Given value of n` `    ``const` `int` `n = 100;`   `    ``// Initialize the sum` `    ``int` `sum = 0;`   `    ``// Loop through odd numbers from 1 to 2*n with step 2` `    ``for` `(``int` `i = 1; i < 2 * n; i += 2) {` `        ``// Add the current odd number to the sum` `        ``sum += i;` `    ``}`   `    ``// Print the sum of the first n terms of the sequence` `    ``cout << ``"Sum of first "` `<< n << ``" terms of the sequence is: "` `<< sum <

## Java

 `public` `class` `SumOfOddNumbers {` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``// Given value of n` `        ``final` `int` `n = ``100``;`   `        ``// Initialize the sum` `        ``int` `sum = ``0``;`   `        ``// Loop through odd numbers from 1 to 2*n with step` `        ``// 2` `        ``for` `(``int` `i = ``1``; i < ``2` `* n; i += ``2``) {` `            ``// Add the current odd number to the sum` `            ``sum += i;` `        ``}`   `        ``// Print the sum of the first n terms of the` `        ``// sequence` `        ``System.out.println(``"Sum of first "` `+ n` `                           ``+ ``" terms of the sequence is: "` `                           ``+ sum);` `    ``}` `}`

## Python3

 `n ``=` `100` `sum` `=` `0` `for` `i ``in` `range``(``1``, ``2``*``n, ``2``):` `    ``sum` `+``=` `i` `print``(``"Sum of first {} terms of the sequence is: {}"``.``format``(n, ``sum``))`

## C#

 `using` `System;`   `class` `Program {` `    ``static` `void` `Main()` `    ``{` `        ``// Given value of n` `        ``const` `int` `n = 100;`   `        ``// Initialize the sum` `        ``int` `sum = 0;`   `        ``// Loop through odd numbers from 1 to 2*n with step` `        ``// 2` `        ``for` `(``int` `i = 1; i < 2 * n; i += 2) {` `            ``// Add the current odd number to the sum` `            ``sum += i;` `        ``}`   `        ``// Print the sum of the first n terms of the` `        ``// sequence` `        ``Console.WriteLine(``"Sum of first "` `+ n` `                          ``+ ``" terms of the sequence is: "` `                          ``+ sum);` `    ``}` `}`

## Javascript

 `// Given value of n` `const n = 100;`   `// Initialize the sum` `let sum = 0;`   `// Loop through odd numbers from 1 to 2*n with step 2` `for` `(let i = 1; i < 2 * n; i += 2) {` `    ``// Add the current odd number to the sum` `    ``sum += i;` `}`   `// Print the sum of the first n terms of the sequence` `console.log(`Sum of first \${n} terms of the sequence is: \${sum}`);`

Output

```Sum of first 100 terms of the sequence is: 10000

```

Time Complexity: O(n) The for loop iterates n times.
Space Complexity: O(1) Only a constant amount of extra space is required to store the sum and the loop variable ‘i’.

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