Question

If a sequence of terms \(u_{n}\) satisfies the recurrence relation \(u_{n+1}=(1-x) u_{n}+n x\) with \(u_{1}=0\) then show, using induction, that for \(n \geq 1\) $$ u_{n}=\frac{1}{x}\left[n x-1+(1-x)^{n}\right] $$

Short answer

Base case and inductive step are verified; formula is true for all \(n \geq 1\).

Step-by-step solution

Base Case

Verify the base case for the given sequence. Given: \(u_1 = 0\)We need to check if the formula holds for \(n = 1\). The formula is \(u_n = \frac{1}{x}\left[nx - 1 + (1 - x)^n\right]\). Substitute \(n = 1\): \(u_1 = \frac{1}{x}\left[1x - 1 + (1 - x)^1\right] = \frac{1}{x}\left[x - 1 + (1 - x)\right] = \frac{1}{x}\left[x - 1 + 1 - x\right] = \frac{1}{x} \cdot 0 = 0\)Therefore, the base case holds true.

Induction Hypothesis

Assume that the formula holds for some integer \(k\), i.e., \(u_k = \frac{1}{x}\left[kx - 1 + (1 - x)^k\right]\).This is our induction hypothesis.

Inductive Step

We need to show that if the formula holds for \(n = k\), then it also holds for \(n = k + 1\). Starting from the recurrence relation: \(u_{k+1} = (1 - x)u_k + kx\).Substitute the induction hypothesis: \(u_{k+1} = (1 - x) \left(\frac{1}{x}[kx - 1 + (1 - x)^k]\right) + kx\).Simplify inside the brackets: \(u_{k+1} = \frac{1}{x}\left[(1 - x)(kx - 1 + (1 - x)^k)\right] + kx\).\(u_{k+1} = \frac{1}{x}[(1 - x)kx - (1 - x) + (1 - x)^{k+1}] + kx\).Distribute \((1 - x)\): \(u_{k+1} = \frac{1}{x}[(1 - x)kx - (1 - x) + (1 - x)^{k+1}] + kx\).Combine terms: \(u_{k+1} = \frac{(1 - x)kx}{x} - \frac{1 - x}{x} + \frac{(1 - x)^{k+1}}{x} + kx\) can be further simplified to\(u_{k+1} = k(1-x) + (1 - x)^{k+1}x^{-1} - (1-x)x^{-1} + kx\) .Combine similar terms:\(u_{k+1} = \frac{kx - (1-x) + (1-x)^{k+1}}{x}\)

Conclusion

Thus the induction step holds, therefore by the principle of mathematical induction, the formula \(u_n = \frac{1}{x}[nx - 1 + (1 - x)^n]\) holds for all \(n \geq 1\).

Key concepts

Recurrence Relation

A recurrence relation is an equation that defines a sequence where each term is a function of its preceding terms. In simpler terms, it tells you how to find the next term in the sequence from the current one.
For our exercise, the recurrence relation is given by:

\[u_{n+1} = (1-x)u_n + nx\]

This equation shows how each term, \(u_{n+1}\), can be found using the previous term, \(u_n\), and some additional operations involving the variable \(n\) and the factor \(x\).
Understanding and solving recurrence relations is crucial in many areas of mathematics and computer science because they often describe complex systems in a simplified, recursive way.

Sequence of Terms

The sequence of terms is simply a list of numbers generated according to some rule or function. In this case, we have a sequence \(u_n\), where each term is generated using the recurrence relation mentioned above.
The given sequence starts with:\

• \(u_1 = 0\)

Using the recurrence relation, each subsequent term is calculated. This creates a sequence of values for \(u_2, u_3, u_4\), and so on.
Sequences are foundational in mathematics and help us understand patterns and behaviors of numerical lists. They appear in various fields such as finance, computer algorithms, and even biology.

Inductive Step

The inductive step is a crucial part of mathematical induction. It involves proving that if a statement holds true for an arbitrary case, say \(n = k\), then it must also be true for the next case, \(n = k + 1\).
In our exercise, we assume the formula holds for \(n = k\), that is:

\[u_k = \frac{1}{x}[kx - 1 + (1 - x)^k]\]
Next, we need to show it holds for \(n = k + 1\). This involves some algebraic manipulation based on the given recurrence relation:
\[u_{k+1} = (1 - x)u_k + kx\]
….Through understanding and simplifying, we show:
\[u_{k+1} = \frac{kx - 1 + (1-x)^{k+1}}{x}\]
The successful completion of the inductive step ensures that our formula is valid for all terms in the sequence.

Base Case

The base case in mathematical induction is the starting point of the proof. It establishes the validity of the statement for the initial value. In our problem, the base case is for \(n = 1\).
We need to verify that the formula:
\[u_n = \frac{1}{x}[nx - 1 + (1 - x)^n]\]\holds for \(n = 1\).
Substituting \(n = 1\) gives:

\[u_1 = \frac{1}{x}[1x - 1 + (1 - x)^1] = \frac{1}{x}[x - 1 + (1 - x)] = \frac{1}{x}[0] = 0\]

As shown, the formula holds true for \(n = 1\), confirming our base case.
Proving the base case is essential as it provides the foundation for the inductive step, ensuring the logical consistency of the entire proof.