# Find sum of series 1^3+1^2+1+2^3+2^2+2+3^3+3^2+3+… till 3N terms

• Last Updated : 16 Aug, 2022

Given a number N, the task is to find the sum of the below series till 3N terms.

1^3+1^2+1+2^3+2^2+2+3^3+3^2+3+… till 3N terms

Examples:

Input: N = 2
Output: 17

Input: N = 3
Output: 56

Naive Approach:

If we observe clearly then we can divide it into a grouping of 3 terms having N no. of groups.

1 to 3 term = 1^3 +1^2 +1 = 3

4 to 6 term = 2^3+2^2+2 = 14

7 to 9 term = 3^3+3^2+ 3 = 39

.

.

(3N-2) to 3N term = N^3+N^2+ N

Below steps can be used to solve the problem-

• For each iterative i, calculate (i^3+i^2+i).
• And add the calculated value to sum (Initially the sum will be 0).
• Return the final sum.

Below is the implementation of the above approach:

## C++

 // C++ program to find the sum of the // series 1^3+1^2+1+2^3+2^2+2+3^3+3^2+3+... // till 3N terms   #include  using namespace std;   // Function to return the sum // upto 3Nth term of the series int seriesSum(int N) {     // Initial value of the sum     int sum = 0;       // Loop to iterate from 1 to N     for (int i = 1; i <= N; i++)      {         // Adding current calculated value          // to sum         sum += (pow(i, 3) + pow(i, 2) + i);     }       // Return the sum upto 3Nth term     return sum; }   // Driver Code int main() {     // Get the value of N     int N = 5;     cout << seriesSum(N);     return 0; }

## Java

 // Java program for the above approach import java.io.*; import java.lang.*; import java.util.*;   class GFG {     // Function to return the sum   // upto 3Nth term of the series   static int seriesSum(int N)   {           // Initial value of the sum     int sum = 0;       // Loop to iterate from 1 to N     for (int i = 1; i <= N; i++)      {               // Adding current calculated value        // to sum       sum += (Math.pow(i, 3) + Math.pow(i, 2) + i);     }       // Return the sum upto 3Nth term     return sum;   }     // Driver Code   public static void main (String[] args)   {     int N = 5;     System.out.print(seriesSum(N));   } }   // This code is contributed by hrithikgarg03188

## Python3

 # Python code for the above approach    # Function to return the sum # upto 3Nth term of the series def seriesSum(N):         # Initial value of the sum     sum = 0;       # Loop to iterate from 1 to N     for i in range(1, N + 1):         # Adding current calculated value          # to sum         sum += (i ** 3) + (i ** 2) + i;       # Return the sum upto 3Nth term     return sum;   # Driver Code   # Get the value of N N = 5; print(seriesSum(N));   # This code is contributed by Saurabh Jaiswal

## C#

 // C# program for the above approach using System; class GFG {     // Function to return the sum   // upto 3Nth term of the series   static int seriesSum(int N)   {           // Initial value of the sum     int sum = 0;       // Loop to iterate from 1 to N     for (int i = 1; i <= N; i++)      {               // Adding current calculated value        // to sum       sum += ((int)Math.Pow(i, 3) + (int)Math.Pow(i, 2) + i);     }       // Return the sum upto 3Nth term     return sum;   }     // Driver Code   public static void Main ()   {     int N = 5;     Console.Write(seriesSum(N));   } }   // This code is contributed by Samim Hossain Mondal.

## Javascript

 

Output

295

Time Complexity: O(N)

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

Efficient Approach:

From the given series, find the formula for the 3Nth term:

The given Series

This can be written as-

-(1)

The above three equations are in A.P., hence can be written as-

= N*(N+1)*(3*N^2+7*N+8)/12

So, the sum of the series till 3Nth term can be generalized as:

## C++

 // C++ program to find the sum of the // series 1^3+1^2+1+2^3+2^2+2+3^3+3^2+3+... // till 3N terms   #include  using namespace std;   // Function to return the sum // upto 3Nth term of the series int seriesSum(int N) {     return N * (N + 1) * (3 * pow(N, 2) + 7 * N + 8) / 12; }   // Driver Code int main() {     // Get the value of N     int N = 5;     cout << seriesSum(N);     return 0; }

## Java

 // Java program to find the sum of the // series 1^3+1^2+1+2^3+2^2+2+3^3+3^2+3+... // till 3N terms import java.util.*; public class GFG {     // Function to return the sum   // upto 3Nth term of the series   static int seriesSum(int N)   {     return N * (N + 1) * (3 * (int)Math.pow(N, 2) + 7 * N + 8) / 12;   }     // Driver Code   public static void main(String args[])   {       // Get the value of N     int N = 5;     System.out.print(seriesSum(N));   } }   // This code is contributed by Samim Hosdsain Mondal.

## Python

 # Pyhton program to find the sum of the # series 1^3+1^2+1+2^3+2^2+2+3^3+3^2+3+... # till 3N terms import math   # Function to return the sum # upto 3Nth term of the series def seriesSum(N):       return math.floor(N * (N + 1) * (3 * pow(N, 2) + 7 * N + 8) / 12)   # Driver Code   # Get the value of N N = 5 print(seriesSum(N))   # This code is contributed by Samim Hossain Mondal

## C#

 // C# program to find the sum of the // series 1^3+1^2+1+2^3+2^2+2+3^3+3^2+3+... // till 3N terms using System; class GFG {     // Function to return the sum // upto 3Nth term of the series static int seriesSum(int N) {     return N * (N + 1) * (3 * (int)Math.Pow(N, 2) + 7 * N + 8) / 12; }   // Driver Code public static void Main() {         // Get the value of N     int N = 5;     Console.Write(seriesSum(N)); } }   // This code is contributed by Samim Hosdsain Mondal.

## Javascript

 

Output

295

Time Complexity: O(1)

Auxiliary Space: O(1)

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