Any cube, n³, can be expressed as the sum of 'n' consecutive odd numbers.
Understanding the Pattern
The key is to find the starting odd number for the sequence that sums to n³. Let's illustrate the general pattern:
- 1³ = 1
- 2³ = 3 + 5
- 3³ = 7 + 9 + 11
- 4³ = 13 + 15 + 17 + 19
- 5³ = 21 + 23 + 25 + 27 + 29
Notice that each cube is represented by a sum of n consecutive odd integers.
Deriving the Formula
The sum of n consecutive odd numbers starting from 'a' can be represented as:
a + (a + 2) + (a + 4) + ... + (a + 2(n-1))
This is an arithmetic progression with n terms, first term 'a', and a common difference of 2. The sum (S) of an arithmetic progression is:
S = (n/2) [2a + (n - 1) d]
In our case, S = n³ and d = 2. So:
n³ = (n/2) [2a + (n - 1) 2]
n³ = (n/2) [2a + 2n - 2]
n³ = n [a + n - 1]
Dividing both sides by n:
n² = a + n - 1
Therefore, the first odd number (a) in the sequence is:
a = n² - n + 1
Applying the Formula
Let's verify this formula with an example. Let's take n = 4:
a = 4² - 4 + 1 = 16 - 4 + 1 = 13
So, the sum of 4 consecutive odd numbers starting from 13 should be equal to 4³ (which is 64).
13 + 15 + 17 + 19 = 64
This confirms our formula.
General Expression
To express any n³ as the sum of consecutive odd numbers:
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Calculate the first odd number: a = n² - n + 1
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Write the sum of 'n' consecutive odd numbers starting from 'a':
n³ = (n² - n + 1) + (n² - n + 3) + (n² - n + 5) + ... + (n² - n + 2(n-1))
or
n³ = Σ [n² - n + (2k-1)] where k goes from 1 to n
Example: Expressing 6³ as a Sum of Consecutive Odd Numbers
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Calculate the first odd number (a):
a = 6² - 6 + 1 = 36 - 6 + 1 = 31
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Write the sum of 6 consecutive odd numbers starting from 31:
6³ = 31 + 33 + 35 + 37 + 39 + 41 = 216
Summary
The formula a = n² - n + 1 allows you to find the starting odd number 'a' required to express any cube (n³) as the sum of 'n' consecutive odd numbers.
