Discrete calculus and sums of powers

Discrete calculus provides a powerful set of tools for analysing sequences and summing polynomials. The backward difference operator $\nabla$ is the discrete counterpart of the derivative, and in this problem we see how repeated differences, together with telescoping sums, unlock closed formulas for sums of powers like $\sum n^2$. By working through these steps, you’ll strengthen your understanding of a major syllabus theme—sequences and series—and gain a technique that generalises to sums of any polynomial degree.

Problem Statement

$\nabla$ is the backward difference operator, defined as $\nabla f(n)=f(n)-f(n-1)$ for discrete functions $f(n)$. In this problem, you will explore the properties of the backward difference operator and its applications to sums of powers.

For example, The first order of $\nabla$ is

$\nabla(n^2)=n^2-(n-1)^2=2n-1$.

The second order of $\nabla$ is

$$\nabla^2 f(n)=\nabla(\nabla f(n))=\nabla(f(n)-f(n-1))=f(n)-2f(n-1)+f(n-2)$$

and the third order of $\nabla$ is $\nabla^3 f(n)=\nabla(\nabla^2 f(n))=\nabla(f(n)-2f(n-1)+f(n-2))=f(n)-3f(n-1)+3f(n-2)-f(n-3)$.

In this problem, we use the $\nabla$ operator to derive formulas for sums of powers.

• Show $n^3-(n-1)^3=3n^2-3n+1$.
• Sum this identity from $1$ to $N$ to derive

$$\sum_{n=1}^N n^2=\frac{N(N+1)(2N+1)}{6}.$$

• Evaluate $\nabla^3(n^3)$

% where $\nabla f(n)=f(n)-f(n-1)$ is the backward difference operator % $\nabla^3 = \nabla \circ \nabla \circ \nabla$ is the third-order backward difference operator, % e.g. $\nabla^3(n^3) = n^3 - 3(n-1)^3 + 3(n-2)^2 - (n-3)^3$.

• State the leading term of $\sum_{n=1}^N n^k$.

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Hints

Use telescoping on the left and known formulas for $\sum n$.

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Solutions

To find the sum of squares, we first need the cubic difference identity. Expanding $(n-1)^3$ and subtracting from $n^3$ gives part (i). Next, we sum this identity from $n=1$ to $N$; on the left, the sum $n^3-(n-1)^3$ telescopes, leaving only $N^3 - 0^3 = N^3$:

$$N^3=3\sum_{n=1}^N n^2-3\sum_{n=1}^N n+N,$$

Since we know $\sum_{n=1}^N n = \frac{N(N+1)}{2}$, we can substitute and solve for the sum of squares. The equation simplifies to

$$\sum_{n=1}^N n^2=\frac{N(N+1)(2N+1)}{6}.$$

For the third-order backward difference, we apply the definition three times (or expand using the pattern seen earlier). Evaluating $\nabla^3(n^3)$ yields the constant $6$. This constant is key: it tells us the leading coefficient for the sum of cubes.

Also $\nabla^3(n^3)=6$. In general, analysing the $k$th order backward difference of $n^k$ reveals that $\nabla^k (n^k) = k!$, which forces the leading term of $\sum_{n=1}^N n^k$ to be $\frac{1}{k+1} N^{k+1}$. Hence,

$$\sum_{n=1}^N n^k=\frac{1}{k+1}N^{k+1}+\text{(lower powers of }N\text{)}.$$

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Takeaways

• The backward difference operator $\nabla$ is a discrete derivative; repeated differences of a polynomial eventually become constant, just as repeated derivatives do.
• Combining a telescoping sum with known lower-order sums lets you derive formulas for sums of squares, cubes, and beyond.
• The $k$th backward difference of $n^k$ equals $k!$, so the leading term of $\sum n^k$ is always $\frac{1}{k+1} N^{k+1}$.

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Further Readings

HSC Sequences, HSC Functions, HSC Integrals, HSC Complex Numbers