---
title: "z-transform of n(n-1)"
subject: "Applied Mathematics"
chapter: 2
key: "semester_4/applied_mathematics:z-transform-n-n-minus-1"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/z-transform-n-n-minus-1
site: BCT Engineering
---

# z-transform of n(n-1)

Applied Mathematics · Chapter 2: The Z-Transform

- Asked in: 2069 Bhadra
- Video: [Z-Transform of (n+1)^2 | Important VTU Problem](https://www.youtube.com/watch?v=7D4fuPbgKkQ) — Mathematics Tutor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/z-transform-n-n-minus-1
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-2

## Question

Find the z-transform of n(n - 1); n = k.

## Short answer

Using Σ k(k-1) w^k = 2w^2/(1-w)^3 with w=1/z yields Z[n(n-1)] = 2z/(z-1)^3, valid for |z|>1.

## Answer

**Meaning.** The sequence is x(k) = k(k − 1), k ≥ 0 (n = k). Its z-transform is X(z) = Σ k(k − 1) z⁻ᵏ.

**Method (geometric series).** For |w| < 1:

Σ_{k=0}^{∞} wᵏ = 1/(1 − w)

Differentiate twice with respect to w and multiply by w²:

Σ k(k − 1) wᵏ = 2w² / (1 − w)³

Put w = 1/z:

```text
X(z) = 2 z⁻² / (1 − 1/z)³
     = 2 z⁻² · z³ / (z − 1)³
     = 2z / (z − 1)³ ,   |z| > 1
```

**Check by the standard results.** Z[k²] = z(z + 1)/(z − 1)³ and Z[k] = z/(z − 1)². So

Z[k² − k] = [z(z + 1) − z(z − 1)] / (z − 1)³ = 2z/(z − 1)³ ✓

**Answer:** Z[n(n − 1)] = 2z/(z − 1)³.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How does the answer get k(k-1) from the geometric series?
   A. Differentiate sum w^k twice and multiply by w^2
   B. Differentiate once and multiply by w
   C. Integrate sum w^k twice
   D. Square the geometric series
   Answer: A. Second derivative of w^k is k(k-1)w^(k-2); multiplying by w^2 restores w^k. (Key point: Method)

2. For |w| < 1, sum k(k-1) w^k equals
   A. w/(1-w)^2
   B. 2w^2/(1-w)^3
   C. 2w/(1-w)^3
   D. 2w^2/(1-w)^2
   Answer: B. Second derivative of 1/(1-w) is 2/(1-w)^3. (Key point: Series result)

3. Putting w = 1/z gives Z[n(n-1)] =
   A. 2z^2/(z-1)^3
   B. 2/(z-1)^3
   C. 2z/(z-1)^3, |z| > 1
   D. z(z+1)/(z-1)^3
   Answer: C. 2z^-2 z^3/(z-1)^3 = 2z/(z-1)^3. (Key point: Substitution)

4. How is the result verified?
   A. Z[k^2] + Z[k]
   B. Z[k^2] * Z[k]
   C. Z[k] differentiated once
   D. Z[k^2] - Z[k] = z(z+1)/(z-1)^3 - z/(z-1)^2 = 2z/(z-1)^3
   Answer: D. k(k-1) = k^2 - k, and the difference is 2z/(z-1)^3. (Key point: Cross-check)