The Method of Differences — A Clean Proof of the Sum of Cubes

The method of differences is a remarkably elegant tool for evaluating finite sums. When each term of a series can be written in the form f(r+1) − f(r), the sum “collapses” — all interior terms cancel, leaving only a boundary expression. This behaviour is called a telescoping sum.

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1) Telescoping Sums

Assume the general term ur can be written as:

ur = f(r+1) − f(r).

Then the finite sum from r = 1 to r = n becomes:

Σ ur = Σ ( f(r+1) − f(r) ).

To see what happens, write out a few terms:

u₁ = f(2) − f(1)
u₂ = f(3) − f(2)
u₃ = f(4) − f(3)
⋮
uₙ = f(n+1) − f(n)

When these are added, everything cancels except the first and last pieces:

Σ ur = f(n+1) − f(1).

This is the essence of the method: interior structure disappears, leaving just the difference between the final and initial states.

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2) A Classic Application — The Sum of Cubes

We will use this technique to prove the well-known identity:

Σ r³  (r = 1 → n) = ¼ n² (n+1)².

This says: the sum of the first n cubes is the square of the sum 1 + 2 + … + n. One of the most beautiful algebraic formulas in elementary mathematics:

(1 + 2 + … + n)² = ¼ n² (n+1)².

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3) Constructing the Telescoping Term

We want to express r³ as a difference f(r+1) − f(r). Define

f(r) = (r − 1)² · r².

Then:

f(r+1) = r² (r+1)².

Compute the difference:

f(r+1) − f(r) = r²(r+1)² − (r−1)²r².

Factor out r²:

= r² [ (r+1)² − (r−1)² ].

Expand the inner square terms:

(r+1)² = r² + 2r + 1
(r−1)² = r² − 2r + 1

Subtracting gives:

(r+1)² − (r−1)² = 4r.

Therefore:

f(r+1) − f(r) = r² · (4r) = 4r³,

so

r³ = ¼ ( f(r+1) − f(r) ).

This is exactly the form we need.

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4) Apply the Method of Differences

Sum both sides from r = 1 to r = n:

Σ r³ = ¼ Σ ( f(r+1) − f(r) ).

The telescoping property gives:

Σ ( f(r+1) − f(r) ) = f(n+1) − f(1).

So:

Σ r³ = ¼ [ f(n+1) − f(1) ].

Now evaluate the endpoints. From our definition:

f(n+1) = n² (n+1)²
f(1)   = 0² · 1² = 0

Thus:

Σ r³ = ¼ n² (n+1)².

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5) Final Identity

Σ r³ (r = 1 → n) = ¼ n² (n+1)².

Since

1 + 2 + … + n = n(n+1)/2,

we have

Σ r³ = [ n(n+1)/2 ]².

The sum of the first n cubes is the square of the nth triangular number — a stunning symmetry linking linear, quadratic, and cubic sequences.

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Why This Technique Matters

The method of differences provides:

• a fast way to evaluate many finite sums
• a structured approach to identifying cancellations
• a deeper understanding of how algebraic expressions encode pattern

It also highlights a key mathematical strategy: rewrite a problem so that structure becomes visible. Once expressed in the right form, the sum simply takes care of itself.

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Summary

• If a term can be written as f(r+1) − f(r), the sum telescopes.
• Almost all terms vanish, leaving f(n+1) − f(1).
• Choosing f(r) = (r−1)²r² allows us to evaluate Σ r³ immediately.
• The result is the elegant identity Σ r³ = (1 + 2 + … + n)².

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Notes prepared for personal study.