Question 9 of 12

Question 9

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. (a) $a_{n+1}=5a_n^2$, $a_0=2$, $a_n>0$. (b) $a_{n+1}-2a_n=2^n$, $a_0=1$.

Find. A closed-form formula for $a_n$ in each case.

Approach. (a) is nonlinear but becomes linear after a logarithmic substitution $b_n=\log_5 a_n$; (b) is linear first-order with a "resonant" forcing term (same base as the homogeneous solution), so use an $n\cdot2^n$ particular-solution ansatz.

1. a) Linearize by taking $\log_5$. Let $b_n=\log_5 a_n$ (valid since $a_n>0$). From $a_{n+1}=5a_n^2$: $\log_5 a_{n+1}=\log_5 5+2\log_5 a_n$, i.e. $b_{n+1}=2b_n+1$, with $b_0=\log_5 2$.
2. a) Solve the linear recurrence $b_{n+1}=2b_n+1$. Adding 1 to both sides: $b_{n+1}+1=2(b_n+1)$, so $b_n+1=2^n(b_0+1)$, i.e. $b_n=2^n(\log_5 2+1)-1$.
3. a) Convert back to $a_n=5^{b_n}$. $$a_n=5^{\,2^n(\log_5 2+1)-1}=\frac15\cdot 5^{2^n(\log_5 2+1)}=\frac15\Big(5^{\log_5 2+1}\Big)^{2^n}=\frac15(2\cdot5)^{2^n}=\frac{10^{2^n}}{5}$$ Check: $a_0=10^1/5=2$ ✓; $a_1=5(2)^2=20$, formula $10^2/5=20$ ✓; $a_2=5(20)^2=2000$, formula $10^4/5=2000$ ✓. $\boxed{a_n=\dfrac{10^{2^n}}{5}}$
4. b) Solve the homogeneous part. $a_{n+1}-2a_n=0$ has solution $a_n^{(h)}=C\cdot2^n$.
5. b) Find a particular solution. Since the forcing term $2^n$ matches the homogeneous root, try $a_n^{(p)}=A\,n\,2^n$. Substituting into $a_{n+1}-2a_n=2^n$: $A(n+1)2^{n+1}-2An2^n=2^n \Rightarrow 2A(n+1)2^n-2An2^n=2^n\Rightarrow 2A\cdot2^n=2^n\Rightarrow A=\tfrac12$. So $a_n^{(p)}=\tfrac{n}{2}2^n=n2^{n-1}$.
6. b) Combine and apply the initial condition. General solution $a_n=C2^n+n2^{n-1}$. At $n=0$: $a_0=C\cdot1+0=C=1$, so $C=1$: $$a_n=2^n+n2^{n-1}=(n+2)2^{n-1}$$ Check: $a_0=2\cdot2^{-1}=1$ ✓; $a_1=3\cdot1=3$, and $2a_0+2^0=2+1=3$ ✓; $a_2=4\cdot2=8$, and $2a_1+2^1=6+2=8$ ✓. $\boxed{a_n=(n+2)2^{n-1}}$