Sum of Cubes of First n Natural Numbers

Given an integer n, Find the sum of series 1³ + 2³ + 3³ + 4³ + ... + n³ till n-th term.

Examples:

Input: n = 5
Output: 225
Explanation: 1³ + 2³ + 3³ + 4³ + 5³ = 225

Input: n = 7
Output: 784
Explanation: 1³ + 2³ + 3³ + 4³ + 5³ + 6³ + 7³ = 784

[Naive Approach] Iterative Method - O(n) Time and O(1) Space

Iterate from 1 to n and keep adding the cube of each number to a running sum. Finally, return the accumulated sum.

Dry run for n=5:

• Initialize sum = 0
• For i = 1, add (1^3 = 1), sum becomes 1
• For i = 2, add (2^3 = 8), sum becomes 9
• For i = 3, add (3^3 = 27), sum becomes 36
• For i = 4, add (4^3 = 64), sum becomes 100
• For i = 5, add (5^3 = 125), sum becomes 225

Final result = 225

225

[Expected Approach] Formula Based Method- O(1) Time and O(1) Space

Use the direct formula (n(n+1)/2)^2 to compute the sum without iteration.

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

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

225

Note: The expression (n*(n+1)/2)^2 may overflow before division for large n.

Rearrange the computation by performing division before multiplication to reduce intermediate values and minimize the risk of integer overflow.

Output: 

225

Time complexity: O(1)
Auxiliary Space: O(1)