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

• Last Updated : 01 Feb, 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

.

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(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 seriesint 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 Codeint main(){    // Get the value of N    int N = 5;    cout << seriesSum(N);    return 0;}

## Java

 // Java program for the above approachimport 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 seriesdef 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 NN = 5;print(seriesSum(N)); # This code is contributed by Saurabh Jaiswal

## C#

 // C# program for the above approachusing 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)

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 seriesint seriesSum(int N){    return N * (N + 1) * (3 * pow(N, 2) + 7 * N + 8) / 12;} // Driver Codeint 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 termsimport 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 termsimport math # Function to return the sum# upto 3Nth term of the seriesdef seriesSum(N):     return math.floor(N * (N + 1) * (3 * pow(N, 2) + 7 * N + 8) / 12) # Driver Code # Get the value of NN = 5print(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 termsusing System;class GFG{   // Function to return the sum// upto 3Nth term of the seriesstatic int seriesSum(int N){    return N * (N + 1) * (3 * (int)Math.Pow(N, 2) + 7 * N + 8) / 12;} // Driver Codepublic 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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