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

• Last Updated : 24 Jan, 2022

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

Input: N = 2
Output: 2.25

Input: N = 3
Output: 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

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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 = 2
Output: 2.25
Explanation: (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 using namespace std; // Function to find N-th term// in the seriesdouble nthTerm(int N){    return (pow(N, 2) +            2 * N + 1) / 4;} // Driver Codeint main(){    // Get the value of N    int N = 5;    cout << nthTerm(N);    return 0;}

## Java

 // Java code for the above approachimport 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 approachimport sys # Function to find N-th term# in the seriesdef nthTerm(N):     return (pow(N, 2) + 2 * N + 1) / 4 # Driver Codeif __name__ == "__main__":            N = 5     print(nthTerm(N))     # This code is contributed by hrithikgarg03188

## C#

 // C# program to find N-th term// in the seriesusing 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

 
Output
9

Time Complexity: O(1)

Auxiliary Space: O(1)

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