C Program for cube sum of first n natural numbers
Print the sum of series 13 + 23 + 33 + 43 + …….+ n3 till n-th term.
Examples:
Input : n = 5 Output : 225 13 + 23 + 33 + 43 + 53 = 225 Input : n = 7 Output : 784 13 + 23 + 33 + 43 + 53 + 63 + 73 = 784
Naive Approach:
- Define the function sumOfCubes which takes an integer input n.
- Declare and initialize a variable sum to 0.
- Start a loop with i ranging from 1 to n.
- For each iteration of the loop, calculate the cube of i and add it to sum.
- After the loop ends, return the value of sum.
C++
#include <iostream> using namespace std; int sumOfCubes( int n) { int sum = 0; for ( int i = 1; i <= n; i++) { sum += i*i*i; } return sum; } int main() { int n = 5; cout << sumOfCubes(n); return 0; } |
225
The time complexity is O(n), where n is the input value.
The auxiliary space is O(1),
An efficient solution is to use direct mathematical formula which is (n ( n + 1 ) / 2) ^ 2
For n = 5 sum by formula is (5*(5 + 1 ) / 2)) ^ 2 = (5*6/2) ^ 2 = (15) ^ 2 = 225 For n = 7, sum by formula is (7*(7 + 1 ) / 2)) ^ 2 = (7*8/2) ^ 2 = (28) ^ 2 = 784
C++
// A formula based C++ program to find sum // of series with cubes of first n natural // numbers #include <iostream> using namespace std; int sumOfSeries( int n) { int x = (n * (n + 1) / 2); return x * x; } // Driver Function int main() { int n = 5; cout << sumOfSeries(n); return 0; } |
225
Time Complexity : O(1)
Auxiliary Space : O(1) because constant space has been used
How does this formula work? We can prove the formula using mathematical induction. We can easily see that the formula holds true for n = 1 and n = 2. Let this be true for n = k-1.
Let the formula be true for n = k-1. Sum of first (k-1) natural numbers = [((k - 1) * k)/2]2 Sum of first k natural numbers = = Sum of (k-1) numbers + k3 = [((k - 1) * k)/2]2 + k3 = [k2(k2 - 2k + 1) + 4k3]/4 = [k4 + 2k3 + k2]/4 = k2(k2 + 2k + 1)/4 = [k*(k+1)/2]2
The above program causes overflow, even if result is not beyond integer limit. Like previous post, we can avoid overflow upto some extent by doing division first. Please refer complete article on Program for cube sum of first n natural numbers for more details!
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