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# Program to print the sum of the given nth term

• Last Updated : 09 Aug, 2021

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 term 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 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...`

## PHP

 ``

## Javascript

 ``

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

 ``

## Javascript

 ``

Output:

`The sum of n term is: 10000`

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

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