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++ 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 series
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 += (pow(Odd, 3)
- pow(Even, 3));
// Increase the odd and
// even with value 2
Odd += 2;
Even += 2;
}
return Sum;
}
// Driver Code
int main()
{
int N = 10;
cout << findSum(N);
}
// 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
# Python 3 program for the above approach
# Function to calculate the sum
# of first N term
def 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 Code
if __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# 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.
<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 series
function 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 Code
let 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++ 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 series
int findSum(int N)
{
return 4 * pow(N, 3) + 9 * pow(N, 2) + 6 * N;
}
// Driver Code
int main()
{
int N = 10;
cout << findSum(N);
}
// 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
# 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 term
def findSum(N):
return 4 * pow(N, 3) + 9 * pow(N, 2) + 6 * N
# Driver Code
if __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# 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.
<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)