# Find Nth term of the series 1^3/1+(1^3+2^3)/(1+3)+(1^3+2^3+3^3)/(1+3+5)+…

Given a positive integer **N**. The task is to find **Nth** term of the series:

**Examples:**

Input:N = 2Output:2.25

Input:N = 3Output:4

**Approach: **

From the given series, find the formula for **Nth** term:

1st term = 1^3/1 = 1/1 = 1

2nd term = (1^3+2^3)/(1+3) = (1+8)/4 = 9/4 = 2.25

3rd term = (1^3+2^3+3^3)/(1+3+5) = (1+8+27)/9 = 4

4th term = (1^3+2^3+3^3+4^3)/(1+3+5+7) = (1+8+27+64)/16 = 6.25

.

.

Nth term = ((N*(N+1)/2)^2)/(N*(2+(N-1)*2)/2) = (N+1)^2/4 = (N^2+2N+1)/4

**Derivation:**

For the series-

Nth term can be written as-

Here,

and 1+3+5+….+(2*N-1) are in A.P.

Rewriting the above equation using the formula for A.P. as-

The **Nth** term of the given series can be generalized as:

**Illustration:**

Input:N = 2Output:2.25Explanation:(1^3+2^3)/(1+3)

= (1 +8)/4

= 9/4

= 2.25

Below is the implementation of the above problem:

## C++

`// C++ program to find N-th term` `// in the series` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find N-th term` `// in the series` `double` `nthTerm(` `int` `N)` `{` ` ` `return` `(` `pow` `(N, 2) + ` ` ` `2 * N + 1) / 4;` `}` `// Driver Code` `int` `main()` `{` ` ` `// Get the value of N` ` ` `int` `N = 5;` ` ` `cout << nthTerm(N);` ` ` `return` `0;` `}` |

## Java

`// Java code for the above approach` `import` `java.io.*;` `class` `GFG {` ` ` `// Function to find N-th term` ` ` `// in the series` ` ` `static` `double` `nthTerm(` `int` `N)` ` ` `{` ` ` `return` `(Math.pow(N, ` `2` `) + ` `2` `* N + ` `1` `) / ` `4` `;` ` ` `}` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `// Get the value of N` ` ` `int` `N = ` `5` `;` ` ` `System.out.println(nthTerm(N));` ` ` `}` `}` `// This code is contributed by Potta Lokesh` |

## Python

`# python 3 program for the above approach` `import` `sys` `# Function to find N-th term` `# in the series` `def` `nthTerm(N):` ` ` `return` `(` `pow` `(N, ` `2` `) ` `+` `2` `*` `N ` `+` `1` `) ` `/` `4` `# Driver Code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` ` ` `N ` `=` `5` ` ` `print` `(nthTerm(N))` ` ` `# This code is contributed by hrithikgarg03188` |

## C#

`// C# program to find N-th term` `// in the series` `using` `System;` `class` `GFG` `{` ` ` `// Function to find N-th term` ` ` `// in the series` ` ` `static` `double` `nthTerm(` `int` `N)` ` ` `{` ` ` `return` `(Math.Pow(N, 2) + ` ` ` `2 * N + 1) / 4;` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `// Get the value of N` ` ` `int` `N = 5;` ` ` `Console.Write(nthTerm(N));` ` ` `}` `}` `// This code is contributed by Samim Hosdsain Mondal.` |

## Javascript

`<script>` ` ` `// JavaScript program to find N-th term` ` ` `// in the series` ` ` `// Function to find N-th term` ` ` `// in the series` ` ` `const nthTerm = (N) => {` ` ` `return` `(Math.pow(N, 2) +` ` ` `2 * N + 1) / 4;` ` ` `}` ` ` `// Driver Code` ` ` `// Get the value of N` ` ` `let N = 5;` ` ` `document.write(nthTerm(N));` `// This code is contributed by rakeshsahni` `</script>` |

**Output**

9

**Time Complexity:** O(1)

**Auxiliary Space:** O(1)