# Program to get the Sum of series: 1 – x^2/2! + x^4/4! -…. upto nth term

• Difficulty Level : Basic
• Last Updated : 23 Nov, 2021

This is a mathematical series program where a user must enter the number of terms up to which the sum of the series is to be found. Following this, we also need the value of x, which forms the base of the series.

Examples:

```Input : x = 9, n = 10
Output : -5.1463

Input : x = 5, n = 15
Output : 0.2837```

Simple approach :
We use two nested loops to compute factorial and use power function to compute power.

## C

 `// C program to get the sum of the series``#include ``#include ` `// Function to get the series``double` `Series(``double` `x, ``int` `n)``{``    ``double` `sum = 1, term = 1, fct, j, y = 2, m;` `    ``// Sum of n-1 terms starting from 2nd term``    ``int` `i;``    ``for` `(i = 1; i < n; i++) {``        ``fct = 1;``        ``for` `(j = 1; j <= y; j++) {``            ``fct = fct * j;``        ``}``        ``term = term * (-1);``        ``m = term * ``pow``(x, y) / fct;``        ``sum = sum + m;``        ``y += 2;``    ``}``    ``return` `sum;``}` `// Driver Code``int` `main()``{``    ``double` `x = 9;``    ``int` `n = 10;``    ``printf``(``"%.4f"``, Series(x, n));``    ``return` `0;``}`

## Java

 `// Java program to get the sum of the series``import` `java.io.*;` `class` `MathSeries {` `    ``// Function to get the series``    ``static` `double` `Series(``double` `x, ``int` `n)``    ``{``        ``double` `sum = ``1``, term = ``1``, fct, j, y = ``2``, m;` `       ``// Sum of n-1 terms starting from 2nd term``        ``int` `i;``        ``for` `(i = ``1``; i < n; i++) {``            ``fct = ``1``;``            ``for` `(j = ``1``; j <= y; j++) {``                ``fct = fct * j;``            ``}``            ``term = term * (-``1``);``            ``m = Math.pow(x, y) / fct;``            ``m = m * term;``            ``sum = sum + m;``            ``y += ``2``;``        ``}``        ``return` `sum;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``double` `x = ``3``;``        ``int` `n = ``4``;``        ``System.out.println(Math.round(Series(x, n) *``                                ``10000.0``) / ``10000.0``);``    ``}``}`

## Python3

 `# Python3 code to get the sum of the series``import` `math` `# Function to get the series``def` `Series( x , n ):``    ``sum` `=` `1``    ``term ``=` `1``    ``y ``=` `2``    ` `    ``# Sum of n-1 terms starting from 2nd term``    ``for` `i ``in` `range``(``1``,n):``        ``fct ``=` `1``        ``for` `j ``in` `range``(``1``,y``+``1``):``            ``fct ``=` `fct ``*` `j``        ` `        ``term ``=` `term ``*` `(``-``1``)``        ``m ``=` `term ``*` `math.``pow``(x, y) ``/` `fct``        ``sum` `=` `sum` `+` `m``        ``y ``+``=` `2``    ` `    ``return` `sum` `# Driver Code``x ``=` `9``n ``=` `10``print``(``'%.4f'``%` `Series(x, n))` `# This code is contributed by "Sharad_Bhardwaj".`

## C#

 `// C# program to get the sum of the series``using` `System;` `class` `GFG {` `    ``// Function to get the series``    ``static` `double` `Series(``double` `x, ``int` `n)``    ``{``        ``double` `sum = 1, term = 1, fct, j, y = 2, m;` `    ``// Sum of n-1 terms starting from 2nd term``        ``int` `i;``        ``for` `(i = 1; i < n; i++) {``            ``fct = 1;``            ``for` `(j = 1; j <= y; j++) {``                ``fct = fct * j;``            ``}``            ``term = term * (-1);``            ``m = Math.Pow(x, y) / fct;``            ``m = m * term;``            ``sum = sum + m;``            ``y += 2;``        ``}``        ``return` `sum;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``        ``double` `x = 9;``        ``int` `n = 10;``        ``Console.Write(Series(x, n) *``                            ``10000.0 / 10000.0);``    ``}``}` `// This code is contributed by vt_m.`

## PHP

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## Javascript

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## C++

 `// C++ program to get the sum of the series``#include ``using` `namespace` `std;` `// Function to get the series``double` `Series(``double` `x, ``int` `n)``{``    ``double` `sum = 1, term = 1, fct, j, y = 2, m;` `    ``// Sum of n-1 terms starting from 2nd term``    ``int` `i;``    ``for` `(i = 1; i < n; i++) {``        ``fct = 1;``        ``for` `(j = 1; j <= y; j++) {``            ``fct = fct * j;``        ``}``        ``term = term * (-1);``        ``m = term * ``pow``(x, y) / fct;``        ``sum = sum + m;``        ``y += 2;``    ``}``    ``return` `sum;``}` `// Driver Code``int` `main()``{``    ``double` `x = 9;``    ``int` `n = 10;``    ``cout << Series(x, n);``    ``return` `0;``}``// This code is contributed by Samim Hossain Mondal.`
Output
`-5.1463`

Time Complexity: O(n * y)

Auxiliary Space: O(1)

Efficient approach :
We can avoid inner loop and use of power function by using values computed in previous iteration.

## C++

 `// C++ program to get the sum of the series``#include ``#include ` `// Function to get the series``double` `Series(``double` `x, ``int` `n)``{``    ``double` `sum = 1, term = 1, fct = 1, p = 1, multi = 1;``    ` `    ``// Computing sum of remaining n-1 terms.``    ``for` `(``int` `i = 1; i < n; i++) {``        ``fct = fct * multi * (multi+1);``        ``p = p*x*x;``        ``term = (-1) * term;       ``        ``multi += 2;``        ``sum = sum + (term * p)/fct;``    ``}``    ``return` `sum;``}` `// Driver Code``int` `main()``{``    ``double` `x = 9;``    ``int` `n = 10;``    ``printf``(``"%.4f"``, Series(x, n));``    ``return` `0;``}`

## Java

 `// Java program to get``// the sum of the series``import` `java.io.*;` `class` `GFG {``    ` `    ``// Function to get``    ``// the series``    ``static` `double` `Series(``double` `x, ``int` `n)``    ``{``        ``double` `sum = ``1``, term = ``1``, fct = ``1``;``        ``double` `p = ``1``, multi = ``1``;``        ` `        ``// Computing sum of remaining``        ``// n-1 terms.``        ``for` `(``int` `i = ``1``; i < n; i++)``        ``{``            ``fct = fct * multi * (multi + ``1``);``            ``p = p * x * x;``            ``term = (-``1``) * term;    ``            ``multi += ``2``;``            ``sum = sum + (term * p) / fct;``        ``}``        ``return` `sum;``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `main(String args[])``    ``{``        ``double` `x = ``9``;``        ``int` `n = ``10``;``        ``System.out.printf(``"%.4f"``, Series(x, n));``    ``}``}` `// This code is contributed by Nikita Tiwari.`

## Python3

 `# Python3 code to get the sum of the series` `# Function to get the series``def` `Series(x, n):``    ``sum` `=` `1``    ``term ``=` `1``    ``fct ``=` `1``    ``p ``=` `1``    ``multi ``=` `1``    ` `    ``# Computing sum of remaining n-1 terms.``    ``for` `i ``in` `range``(``1``, n):``        ``fct ``=` `fct ``*` `multi ``*` `(multi``+``1``)``        ``p ``=` `p``*``x``*``x``        ``term ``=` `(``-``1``) ``*` `term``        ``multi ``+``=` `2``        ``sum` `=` `sum` `+` `(term ``*` `p)``/``fct``    ` `    ``return` `sum` `# Driver Code``x ``=` `9``n ``=` `10``print``(``'%.4f'``%` `Series(x, n))` `# This code is contributed by "Sharad_Bhardwaj".`

## C#

 `// C# program to get``// the sum of the series``using` `System;` `class` `GFG {``    ` `    ``// Function to get``    ``// the series``    ``static` `float` `Series(``double` `x, ``int` `n)``    ``{``        ``double` `sum = 1, term = 1, fct = 1;``        ``double` `p = 1, multi = 1;``        ` `        ``// Computing sum of remaining``        ``// n-1 terms.``        ``for` `(``int` `i = 1; i < n; i++)``        ``{``            ``fct = fct * multi * (multi + 1);``            ``p = p * x * x;``            ``term = (-1) * term;``            ``multi += 2;``            ``sum = sum + (term * p) / fct;``        ``}``        ``return` `(``float``)sum;``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``        ``double` `x = 9;``        ``int` `n = 10;``        ``Console.Write(Series(x, n));``    ``}``}` `// This code is contributed by vt_m.`

## PHP

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## Javascript

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Output
`-5.1463`

Time Complexity: O(n)

Auxiliary Space: O(1)

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