Print the sum of series 13 + 23 + 33 + 43 + …….+ n3 till n-th term.
Input : n = 5 Output : 225 13 + 23 + 33 + 43 + 53 = 225 Input : n = 7 Output : 784 13 + 23 + 33 + 43 + 53 + 63 + 73 = 784
Time Complexity : O(n)
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
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]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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