Given a positive integer N, the task is to find the sum upto Nth term of the series:
33 – 23, 53 – 43, 73 – 63, …., till N terms
Examples:
Input: N = 10
Output: 4960Input: N = 1
Output: 19
Naive Approach:
- Initialize two int variables odd and even. Odd with value 3 and even with value 2.
- Now Iterate the for loop n times each time will calculate the current term and add it to the sum.
- In each iteration increase odd and even value with 2.
- Return the resultant sum
C++
// C++ program to find sum of N terms of the// series 3^3-2^3, 5^3 - 4^3, 7^3 - 6^3, ...#include <bits/stdc++.h>using namespace std;// Function to return sum of// N term of the seriesint findSum(int N){ // Initialize the variable int Odd = 3; int Even = 2; int Sum = 0; // Run a loop for N number of times for (int i = 0; i < N; i++) { // Calculate the current term // and add it to the sum Sum += (pow(Odd, 3) - pow(Even, 3)); // Increase the odd and // even with value 2 Odd += 2; Even += 2; } return Sum;}// Driver Codeint main(){ int N = 10; cout << findSum(N);} |
Java
// JAVA program to find sum of N terms of the// series 3^3-2^3, 5^3 - 4^3, 7^3 - 6^3, ...import java.util.*;class GFG { // Function to return sum of // N term of the series public static int findSum(int N) { // Initialize the variable int Odd = 3; int Even = 2; int Sum = 0; // Run a loop for N number of times for (int i = 0; i < N; i++) { // Calculate the current term // and add it to the sum Sum += (Math.pow(Odd, 3) - Math.pow(Even, 3)); // Increase the odd and // even with value 2 Odd += 2; Even += 2; } return Sum; } // Driver Code public static void main(String[] args) { int N = 10; System.out.print(findSum(N)); }}// This code is contributed by Taranpreet |
Python3
# Python 3 program for the above approach# Function to calculate the sum# of first N termdef findSum(N): # Initialize the variable Odd = 3 Even = 2 Sum = 0 # Run a loop for N number of times for i in range(N): # Calculate the current term # and add it to the sum Sum += (pow(Odd, 3) - pow(Even, 3)) # Increase the odd and # even with value 2 Odd += 2 Even += 2 return Sum# Driver Codeif __name__ == "__main__": # Value of N N = 10 # Function call to calculate # sum of the series print(findSum(N))# This code is contributed by Abhishek Thakur. |
C#
// C# program to find sum of N terms of the// series 3^3-2^3, 5^3 - 4^3, 7^3 - 6^3, ...using System;class GFG { // Function to return sum of // N term of the series public static int findSum(int N) { // Initialize the variable int Odd = 3; int Even = 2; int Sum = 0; // Run a loop for N number of times for (int i = 0; i < N; i++) { // Calculate the current term // and add it to the sum Sum += (int)(Math. Pow(Odd, 3) - Math.Pow(Even, 3)); // Increase the odd and // even with value 2 Odd += 2; Even += 2; } return Sum; } // Driver Code public static void Main() { int N = 10; Console.Write(findSum(N)); }}// This code is contributed by Samim Hossain Mondal. |
Javascript
<script>// Javascript program to find sum of N terms of the// series 3^3-2^3, 5^3 - 4^3, 7^3 - 6^3, ...// Function to return sum of// N term of the seriesfunction findSum(N) { // Initialize the variable let Odd = 3; let Even = 2; let Sum = 0; // Run a loop for N number of times for (let i = 0; i < N; i++) { // Calculate the current term // and add it to the sum Sum += (Math.pow(Odd, 3) - Math.pow(Even, 3)); // Increase the odd and // even with value 2 Odd += 2; Even += 2; } return Sum;}// Driver Codelet N = 10;document.write(findSum(N));// This code is contributed by gfgking.</script> |
Output
4960
Time Complexity: O(N)
Auxiliary Space: O(1), since no extra space has been taken.
Efficient Approach:
The sequence is formed by using the following pattern.
For any value N the generalise form of the given sequence is-
SN = 4*N3 + 9*N2 + 6*N
Below is the implementation of the above approach:
C++
// C++ program to find the sum of N terms of the// series 3^3-2^3, 5^3 - 4^3, 7^3 - 6^3, ...#include <bits/stdc++.h>using namespace std;// Function to return sum of// N term of the seriesint findSum(int N){ return 4 * pow(N, 3) + 9 * pow(N, 2) + 6 * N;}// Driver Codeint main(){ int N = 10; cout << findSum(N);} |
Java
// Java program to find the sum of N terms of the// series 3^3-2^3, 5^3 - 4^3, 7^3 - 6^3, ...import java.util.*;class GFG{ // Function to return sum of // N term of the series static int findSum(int N) { return (int) (4 * Math.pow(N, 3) + 9 * Math.pow(N, 2) + 6 * N); } // Driver Code public static void main(String[] args) { int N = 10; System.out.print(findSum(N)); }}// This code is contributed by 29AjayKumar |
Python3
# Python 3 program to find the sum of N terms of the# series 3^3-2^3, 5^3 - 4^3, 7^3 - 6^3, ...# Function to calculate the sum# of first N termdef findSum(N): return 4 * pow(N, 3) + 9 * pow(N, 2) + 6 * N# Driver Codeif __name__ == "__main__": # Value of N N = 10 # Function call to calculate # sum of the series print(findSum(N))# This code is contributed by Abhishek Thakur. |
C#
// C# program to find the sum of N terms of the// series 3^3-2^3, 5^3 - 4^3, 7^3 - 6^3, ...using System;class GFG { // Function to return sum of // N term of the series static int findSum(int N) { return 4 * (int)Math.Pow(N, 3) + 9 * (int)Math.Pow(N, 2) + 6 * N; } // Driver Code public static void Main() { int N = 10; Console.Write(findSum(N)); }}// This code is contributed by ukasp. |
Javascript
<script> // JavaScript code for the above approach // Function to return sum of // N term of the series function findSum(N) { return 4 * Math.pow(N, 3) + 9 * Math.pow(N, 2) + 6 * N; } // Driver Code let N = 10; document.write(findSum(N)); // This code is contributed by Potta Lokesh </script> |
Output
4960
Time Complexity: O(1)
Auxiliary Space: O(1)
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