Problem of the Week

Problem 16 Solution

Congratulations to Joseph Braithwaite and Felix Reiter from Bottisham Village College for successfully solving this problem.

$\textbf{Solution:}$

$\textbf{i)}$
$F(F(b)) = a(ab + b) + b = a^2b + ab + b$
$F(F(F(b))) = a(a^2b + ab + b) + b = a^3b + a^2b + ab + b$

$\textbf{ii)}$
$\frac{b}{1-a}$
From part $\textbf{i}$ , we observe:
$F(F(F(\cdots F(b)))) = b + ba + ba^2 + ba^3 + \cdots = b(1 + a + a^2 + a^3 + \cdots)$
(The proof is not needed, this can be assumed based on the pattern observed.)
Note that we accepted answers that had this form.

Let $S = 1 + a + a^2 + a^3 + \cdots$ .
Then:
$aS = a + a^2 + a^3 + a^4 + \cdots$
$aS + 1 = 1 + a + a^2 + a^3 + \cdots$
$aS + 1 = S$
$S = \frac{1}{1-a}, \quad \text{where the infinite series will only converge when } |a| < 1.$

Thus, we conclude:
$F(F(F(\cdots F(b)))) = b(1 + a + a^2 + a^3 + \cdots) = b \cdot \frac{1}{1-a} = \frac{b}{1-a}$

Problem of the Week

Problem 16 Solution

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