The sum of the first n consecutive cubes is equal to the square of the sum of the first n numbers. This post explains how to analyse the pattern of the sum of consecutive cubes and the square of the sum of the first n numbers, derive the sum formula and prove the formula using mathematical duction.
Analysis of cubes
\( \begin{align} 1^3 &= 1 \\ 2^3 &= 2 \times 2^2 \\ &= 2 \times 4 \\ &= (3-1)(3+1) \\ &= 3^2-1^2 \\ 3^3 &= 3 \times 3^2 \\ &= 3 \times 9 \\ &= (6-3)(6+3) \\ &= 6^2-3^2 \\ 4^3 &= 4 \times 4^2 \\ &= 4 \times 16 \\ &= (10-6)(10+6) \\ &= 10^2-6^2 \\ 5^3 &= 5 \times 5^2 \\ &= 5 \times 25 \\ &= (15-10)(15+10) \\ &= 15^2-10^2 \\ &\cdots \end{align} \)
Pattern of cubes
The difference between the squares of two consecutive triangles equals a cube number.
You may notice the following:
\( \begin{align} 2 &= 3-1 \leadsto 2^3 = 3^2-1^2 \\ 3 &= 6-3 \leadsto 3^3 = 6^2-3^2 \\ 4 &= 10-6 \leadsto 4^3 = 10^2-6^2 \\ 5 &= 15-10 \leadsto 5^3 = 15^2-10^2 \\ &\cdots \end{align} \)
Sum of cubes
\( \begin{align} 1^3 &= (1)^2 \\ 1^3 + 2^3 &= 1^2+ (3^2-1^2) \\ &= 3^2 \\ &= (1+2)^2 \\ 1^3 + 2^3 + 3^3 &= 1^2 + (3^2-1^2) + (6^2-3^2) \\ &= 6^2 \\ &= (1+2+3)^2 \\ 1^3 + 2^3 + 3^3 + 4^3 &= 1^2 + (3^2-1^2) + (6^2-3^2) + (10^2-6^2) \\ &= 10^2 \\ &= (1+2+3+4)^2 \\ 1^3 + 2^3 + 3^3 + 4^3 + 5^3 &= 1^2 + (3^2-1^2) + (6^2-3^2) + (10^2-6^2) + (15^2-10^2) \\ &= 15^2 \\ &= (1+2+3+4+5)^2 \\ &\cdots \require{AMSsymbols} \\ \therefore 1^3+2^3+3^3+4^3+5^3+ \cdots + n^3 &= (1+2+3+4+5+ \cdots + n)^2 \end{align} \)
This remarkable formula is called “The sum of \( n \) consecutive cube numbers is equal to the square of the \(n\) numbers”.
Simplification of the formula
This can also be simplified using \( 1+2+3+ \cdots + n = \displaystyle \frac{n(n+1)}{2} \cdots (1) \) .
\( \begin{align} (1+2+3+4+5+ \cdots + n)^2 &= \displaystyle \left[ \frac{n(n+1)}{2} \right]^2 \\ &= \frac{n^2 (n+1)^2}{4} \cdots (2) \end{align} \)
Proving by mathematical induction
Now let’s prove this formula using mathematical induction.
Step 1
Show it is true for \( n=1 \).
LHS \( =1^3=1 \)
RHS \( = 1^2=1 \)
So the formula is true for \( n=1 \).
Step 2
Assume the formula is true for \( n=k \).
That is \( 1^3+2^3+3^3+ \cdots + k^3 = (1+2+3+ \cdots + k)^2 \).
Step 3
Show the formula is true for \( n=k+1 \).
That is \( 1^3+2^3+3^3+ \cdots + k^3 + (k+1)^3 = (1+2+3+ \cdots + k+1)^2 \).
\( \begin{align} \bbox[yellow]{1^3+2^3+3^3+ \cdots + k^3} + (k+1)^3 &= \bbox[yellow]{(1+2+3+ \cdots + k)^2} + (k+1)^3 &\cdots \text{by step 2}\\ &= \displaystyle \frac{k^2 (k+1)^2}{4} + (k+1)^3 &\cdots \text{by } (1) \\ &= \frac{k^2 (k+1)^2}{4} + 4(k+1) \times \frac{(k+1)^2}{4} \\ &= \frac{(k+1)^2}{4} \times (k^2 + 4k+4) &\cdots \text{factorised} \\ &= \frac{(k+1)^2}{4} \times (k+2)^2 \\ &= (1+2+3+ \cdots + k+1)^2 &\cdots \text{by } (2) \end{align} \)
Therefore the formula is true for any integer \( n\ge 1 \).
Frequently Asked Questions
How to find the sum of consecutive cubes?
\( 1^3+2^3+3^3+4^3+5^3+ \cdots + n^3 = (1+2+3+4+5+ \cdots + n)^2 \)
What is the formula for the sum of cubes?
\( \begin{align} 1^3+2^3+3^3+4^3+5^3+ \cdots + n^3 &= (1+2+3+4+5+ \cdots + n)^2 \\ &= \displaystyle \left[ \frac{n(n+1)}{2} \right]^2 \\ &= \frac{n^2 (n+1)^2}{4} \end{align} \)
Is the sum of three consecutive cubes divisible by 9?
\( n^3 + (n+1)^3 + (n+2)^3 \) is divisible by \( 9 \).
Click here to see how to prove this.
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