打印系列1 3 + 2 3 + 3 3 + 4 3 +…。+ n 3的总和直到第n个项。
例子：
Input : n = 5
Output : 225
13 + 23 + 33 + 43 + 53 = 225
Input : n = 7
Output : 784
13 + 23 + 33 + 43 + 53 +
63 + 73 = 784
CPP
// Simple C++ program to find sum of series
// with cubes of first n natural numbers
#include
using namespace std;
/* Returns the sum of series */
int sumOfSeries(int n)
{
int sum = 0;
for (int x=1; x<=n; x++)
sum += x*x*x;
return sum;
}
// Driver Function
int main()
{
int n = 5;
cout << sumOfSeries(n);
return 0;
}
Output :225
Time Complexity : O(n)An efficient solution is to use direct mathematical formula which is (n ( n + 1 ) / 2) ^ 2For 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
Output:225
Time Complexity : O(1)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]2The 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. CPP
// Efficient CPP program to find sum of cubes
// of first n natural numbers that avoids
// overflow if result is going to be withing
// limits.
#include
using namespace std;
// Returns sum of first n natural
// numbers
int sumOfSeries(int n)
{
int x;
if (n % 2 == 0)
x = (n/2) * (n+1);
else
x = ((n + 1) / 2) * n;
return x * x;
}
// Driver code
int main()
{
int n = 5;
cout << sumOfSeries(n);
return 0;
}
Output:225
Please refer complete article on Program for cube sum of first n natural numbers for more details!Want to learn from the best curated videos and practice problems, check out the C++ Foundation Course for Basic to Advanced C++ and C++ STL Course for the language and STL. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.