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# Sum of alternating sign cubes of first N Natural numbers

• Last Updated : 05 Nov, 2021

Given a number N, the task is to find the sum of alternating sign cubes of first N natural numbers, i.e.,

13 – 23 + 33 – 43 + 53 – 63 + ….

Examples:

Input: N = 2
Output: -7
Explanation:
Required sum = 13 – 23 = -7
Input: N = 3
Output: 20
Explanation:
Required sum = 13 – 23 + 33 = 20

Naive Approach: A simple solution is to solve this problem by iterating over a loop from to N and compute the sum by alternating the sign each time.
Below is the implementation of above approach:

## C++

 // C++ implementation to compute // the sum of cubes with // alternating sign   #include    using namespace std;   // Function to compute sum // of the cubes with // alternating sign int summation(int n) {     int sum = 0;     for (int i = 1; i <= n; i++)         if (i % 2 == 1)             sum += (i * i * i);         else             sum -= (i * i * i);       return sum; }   // Driver code int main() {     int n = 3;     cout << summation(n);     return 0; }

## Java

 // Java implementation to compute // the sum of cubes with // alternating sign import java.util.*;   class GFG {   // Function to compute sum // of the cubes with // alternating sign static int summation(int n) {     int sum = 0;           for(int i = 1; i <= n; i++)     {        if (i % 2 == 1)            sum += (i * i * i);        else            sum -= (i * i * i);     }     return sum; }   // Driver code public static void main(String[] args) {     int n = 3;       System.out.println(summation(n)); } }   // This code is contributed by offbeat

## Python3

 # Python3 implementation to   # compute the sum of cubes   # with alternating sign    # Function to compute sum  # of the cubes with  # alternating sign  def summation(n):           sum = 0     for i in range(1, n + 1):         if i % 2 == 1:             sum = sum + (i * i * i)         else:             sum = sum - (i * i * i)       return sum   # Driver code  n = 3   print(summation(n))   # This code is contributed by ishayadav181

## C#

 // C# implementation to compute // the sum of cubes with // alternating sign using System; class GFG{   // Function to compute sum // of the cubes with // alternating sign static int summation(int n) {     int sum = 0;           for(int i = 1; i <= n; i++)     {         if (i % 2 == 1)             sum += (i * i * i);         else             sum -= (i * i * i);     }     return sum; }   // Driver code public static void Main(String[] args) {     int n = 3;       Console.WriteLine(summation(n)); } }   // This code is contributed by sapnasingh4991

## Javascript

 

Output:

20

Time Complexity: O(n)

Auxiliary Space: O(1)

Efficient Approach: The key observation in the problem is that every even number is with a negative sign, that is it used to reduce the overall sum. Therefore if we compute the sum of cubes of even numbers and odd numbers individually, then the overall sum can be computed easily.

• Count of Even or Odd numbers in first N natural numbers
=> => • Sum of first Even Terms
=> • Sum of first Odd Terms
=> • Overall Sum
=> => Below is the implementation of above approach:

## C++

 // C++ implementation to compute // the sum of cubes with // alternating sign   #include  using namespace std;   // Function to compute sum // of the cubes with alternating sign int summation(int N) {     int co = (N + 1) / 2;     int ce = (N) / 2;     int se = 2 * ((ce * (ce + 1))                   * (ce * (ce + 1)));     int so = (co * co)              * (2 * ((co * co)) - 1);     return so - se; }   // Driver Code int main() {     int n = 3;     cout << summation(n);     return 0; }

## Java

 // Java implementation to compute // the sum of cubes with // alternating sign import java.util.*;   class GFG{   // Function to compute sum // of the cubes with // alternating sign static int summation(int N) {     int co = (N + 1) / 2;     int ce = (N) / 2;           int se = 2 * ((ce * (ce + 1)) *                   (ce * (ce + 1)));     int so = (co * co) * (2 * ((co * co)) - 1);           return so - se; }   // Driver code public static void main(String[] args) {     int n = 3;       System.out.println(summation(n)); } }   // This code is contributed by offbeat

## Python3

 # Python3 implementation to compute  # the sum of cubes with  # alternating sign    # Function to compute sum of  # the cubes with alternating sign  def summation(N):           co = (N + 1) / 2     co = int(co)           ce = N / 2     ce = int(ce)           se = 2 * ((ce * (ce + 1)) *               (ce * (ce + 1)))     so = (co * co) * (2 * (co * co) - 1)           return so - se   # Driver Code  n = 3   print(summation(n))   # This code is contributed by ishayadav181

## C#

 // C# implementation to compute // the sum of cubes with // alternating sign using System;   class GFG{   // Function to compute sum // of the cubes with // alternating sign static int summation(int N) {     int co = (N + 1) / 2;     int ce = (N) / 2;           int se = 2 * ((ce * (ce + 1)) *                   (ce * (ce + 1)));     int so = (co * co) * (2 * ((co * co)) - 1);           return so - se; }   // Driver code public static void Main(String[] args) {     int n = 3;       Console.WriteLine(summation(n)); } }   // This code is contributed by Rohit_ranjan

## Javascript

 

Output:

20

Time Complexity: O(1)

Auxiliary Space: O(1)

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