14. What do all linear functions of the form , have in common?

Understand the Question We are given a sequence where the first two terms are and . For any term after the second (), each term is the sum of all preceding terms plus : To solve this problem: 1. Successively compute to by substituting the already calculated values. 2. Compare and to find a simplified recurrence relation relating consecutive terms. 3. Identify the underlying pattern in the sequence terms to write an explicit closed-form formula. Part (i) Find the values of . Step 1 · Calculate Terms Through Given: Using for : Answer > (i) Part (ii) Can you find a simpler recursive formula for ? Step 1 · Derive Simpler Recursive Formula For : For , writing the formula for : Subtracting equation from equation : Checking for and : For : For : Thus, the formula holds for all with initial condition . Answer > (ii) Part (iii) Can you give an explicit formula? Step 1 · Deduce Explicit Formula Express each calculated term as a power of : Answer > (iii) Common Mistakes Power of Two Index Error: Writing instead of . Always verify with the base case . Forgetting Initial Conditions: Stating only without specifying and the domain . Missing the in Summation: Overlooking the constant term at the end of the original summation definition when computing terms manually.

Introduction to Linear Polynomials | EOT

Question 14

*14. What do all linear functions of the form $f(x) = ax + a, a > 0$ , have in common?

Solution

Understand the Question

We are given a sequence where the first two terms are $P_1 = 1$ and $P_2 = 2$ .
For any term after the second ( $n > 2$ ), each term is the sum of all preceding terms plus $1$ : $P_n = P_1 + P_2 + \dots + P_{n-1} + 1$
To solve this problem:
1. Successively compute $P_3$ to $P_8$ by substituting the already calculated values.
2. Compare $P_n$ and $P_{n-1}$ to find a simplified recurrence relation relating consecutive terms.
3. Identify the underlying pattern in the sequence terms to write an explicit closed-form formula.

(i) Find the values of $P_1, P_2, \dots, P_8$ .

Step 1 · Calculate Terms $P_3$ Through $P_8$

Given: $P_1 = 1$ $P_2 = 2$

Using $P_n = P_1 + P_2 + \dots + P_{n-1} + 1$ for $n > 2$ :

\begin{aligned} P_3 &= P_1 + P_2 + 1 \\ &= 1 + 2 + 1 = 4 \end{aligned}

\begin{aligned} P_4 &= P_1 + P_2 + P_3 + 1 \\ &= 1 + 2 + 4 + 1 = 8 \end{aligned}

\begin{aligned} P_5 &= P_1 + P_2 + P_3 + P_4 + 1 \\ &= 1 + 2 + 4 + 8 + 1 = 16 \end{aligned}

\begin{aligned} P_6 &= P_1 + P_2 + P_3 + P_4 + P_5 + 1 \\ &= 1 + 2 + 4 + 8 + 16 + 1 = 32 \end{aligned}

\begin{aligned} P_7 &= P_1 + P_2 + P_3 + P_4 + P_5 + P_6 + 1 \\ &= 1 + 2 + 4 + 8 + 16 + 32 + 1 = 64 \end{aligned}

\begin{aligned} P_8 &= P_1 + P_2 + P_3 + P_4 + P_5 + P_6 + P_7 + 1 \\ &= 1 + 2 + 4 + 8 + 16 + 32 + 64 + 1 = 128 \end{aligned}

Answer

(i) $1, 2, 4, 8, 16, 32, 64, 128$

(ii) Can you find a simpler recursive formula for $P_n$ ?

Step 1 · Derive Simpler Recursive Formula

For $n > 2$ : $P_n = P_1 + P_2 + \dots + P_{n-1} + 1 \quad \dots (1)$

For $n > 3$ , writing the formula for $P_{n-1}$ : $P_{n-1} = P_1 + P_2 + \dots + P_{n-2} + 1 \quad \dots (2)$

Subtracting equation $(2)$ from equation $(1)$ :

\begin{aligned} P_n - P_{n-1} &= (P_1 + \dots + P_{n-1} + 1) - (P_1 + \dots + P_{n-2} + 1) \\ P_n - P_{n-1} &= P_{n-1} \end{aligned}

\begin{aligned} P_n &= P_{n-1} + P_{n-1} \\ &= 2P_{n-1} \end{aligned}

Checking for $n = 2$ and $n = 3$ :

For $n = 2$ : $2P_1 = 2(1) = 2 = P_2$
For $n = 3$ : $2P_2 = 2(2) = 4 = P_3$

Thus, the formula holds for all $n \ge 2$ with initial condition $P_1 = 1$ .

Answer

(ii) $P_1 = 1, \quad P_n = 2P_{n-1} \text{ for } n \ge 2$

(iii) Can you give an explicit formula?

Step 1 · Deduce Explicit Formula

Express each calculated term as a power of $2$ :

\begin{aligned} P_1 &= 1 = 2^0 \\ P_2 &= 2 = 2^1 \\ P_3 &= 4 = 2^2 \\ P_4 &= 8 = 2^3 \\ &\;\vdots \\ P_n &= 2^{n-1} \end{aligned}

Answer

(iii) $P_n = 2^{n-1}$

Common Mistakes

Power of Two Index Error: Writing $P_n = 2^n$ instead of $P_n = 2^{n-1}$ . Always verify with the base case $P_1 = 2^{1-1} = 2^0 = 1$ .
Forgetting Initial Conditions: Stating only $P_n = 2P_{n-1}$ without specifying $P_1 = 1$ and the domain $n \ge 2$ .
Missing the $+1$ in Summation: Overlooking the constant $+1$ term at the end of the original summation definition when computing terms manually.

Question 14

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