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

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  • Last Updated : 16 Aug, 2022
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Given a positive integer N. The task is to find Nth term of the series:

\frac{1^{3}}{1}+\frac{(1^{3}+2^{3})}{1+3}+\frac{(1^{3}+2^{3}+3^{3})}{(1+3+5)}+....

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

.

.

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-

\frac{1^{3}}{1}+\frac{(1^{3}+2^{3})}{1+3}+\frac{(1^{3}+2^{3}+3^{3})}{(1+3+5)}+....

Nth term can be written as-

T_{N}=\frac{(1^{3}+2^{3}+3^{3}+....+N^{3})}{(1+3+5+....+(2*N-1))}

Here,

1^{3}+2^{3}+3^{3}+....+N^{3}              and 1+3+5+….+(2*N-1) are in A.P.

1^{3}+2^{3}+3^{3}+....+N^{3} = (\frac{(N*(N+1))}{2})^{2}

1+3+5+....+(2*N-1)=\frac{(N*(2*1+(N-1)*2))}{2}

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

T_{N}=\frac{(\frac{(N*(N+1))}{2})^{2}}{\frac{(N*(2*1+(N-1)*2))}{2}}

T_{N}=\frac{(N^{2}*(N+1)^{2})}{(2*N*(2+(N-1)*2))}

T_{N}=\frac{(N*(N+1)^{2})}{(4*(1+N-1))}

T_{N}=\frac{(N*(N+1)^{2})}{4*N}

T_{N}=\frac{(N+1)^{2}}{4}

T_{N}=\frac{(N^{2}+2*N+1)}{4}

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

T_{N}=\frac{(N^{2}+2*N+1)}{4}

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 <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)


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