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# Find the Nth term of the series 1 + 2 + 6 + 15 + 31 + 56 + …

Given an integer . The task is to write a program to find the N-th term of the given series:

1 + 2 + 6 + 15 + 31 + 56 + ...

Examples

Input : N = 8
Output : 141

Input : N = 20
Output : 2471

Approach:
The given series is:

1, 2, 6, 15, 31, 56, 92, 141, ...

First consecutive difference

1 4 9 16 25 36 49 .......

Second consecutive difference

3 5 7 9 11 13......

As the second consecutive difference is in AP, the nth term (tn) of the series is of the form,
A(n – 1)(n – 2)(n – 3) + B(n – 1)(n – 2) + C(n – 1) + D
So, tn = A(n – 1)(n – 2)(n – 3) + B(n – 1)(n – 2) + C(n – 1) + D
Now,
t1 = D = 1
t2 = C (2 – 1) + D = 2
t3 = 2B + 2C + D = 6
t4 = CA + 6B + 3C + D = 15
On solving the above four equations we get => A = 1/3, B = 3/2, C = 1, D = 1. On substituting these values tn and after simplifying we get, Below is the implementation of above approach:

## C++

 // C++ program to find Nth  // term of the series: // 1 + 2 + 6 + 15 + 31 + 56 + ... #include #include using namespace std;   // calculate Nth term of given series int Nth_Term(int n) {     return (2 * pow(n, 3) - 3 *                  pow(n, 2) + n + 6) / 6; }   // Driver code int main() {     int N = 8;     cout << Nth_Term(N); }

## Java

 // Java program to find Nth  // term of the series:  // 1 + 2 + 6 + 15 + 31 + 56 + ...  import java.util.*; import java.lang.*;   class GFG { // calculate Nth term of given series  static double Nth_Term(int n)  {      return (2 * Math.pow(n, 3) - 3 *                  Math.pow(n, 2) + n + 6) / 6;  }    // Driver code  static public void main (String args[]) {     int N = 8;      System.out.println(Nth_Term(N));  } }   // This code is contributed // by Akanksha Rai

## Python3

 # Python program to find Nth term of the series: # 1 + 2 + 6 + 15 + 31 + 56 + ...   # calculate Nth term of given series def Nth_Term(n):     return (2 * pow(n, 3) - 3 * pow(n, 2) + n + 6) // 6   # Driver code N = 8 print(Nth_Term(N))

## C#

 // C# program to find Nth  // term of the series:  // 1 + 2 + 6 + 15 + 31 + 56 + ...  using System;   class GFG { // calculate Nth term of given series  static double Nth_Term(int n)  {      return (2 * Math.Pow(n, 3) - 3 *                  Math.Pow(n, 2) + n + 6) / 6;  }    // Driver code  static public void Main () {     int N = 8;      Console.WriteLine(Nth_Term(N));  } }   // This code is contributed // by Sach_Code

## PHP

 

## Javascript

 

Output:

141

Time Complexity: O(1), since there is no loop or recursion.

Auxiliary Space: O(1), since no extra space has been taken.

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