Question 12 Exercise 7.1
Solutions of Question 12 of Exercise 7.1 of Unit 07: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.
Question 12(i)
Show by mathematical induction that $\dfrac{5^{2 n}-1}{24}$ is an integer. Solution: 1. For $n=1$, then $$\dfrac{5^{2 n}-1}{24}=\dfrac{5^{2.1}-1}{24}=\dfrac{24}{24}=1 \in \mathbb{Z}$$ Thus it is true for $n=1$
2. Let it be true for $n=k>1$ then $$\dfrac{5^{2 k}-1}{24} \in \mathbb{Z}$$ 3. For $n=k+1$ then consider \begin{align}\dfrac{5^{2(k+1)}-1}{24}&=\dfrac{5^{2 k+2}-1}{24} \\ & =\dfrac{5^{2 k+2}-1}{24}\\ & =\dfrac{5^{2 k} \cdot 25-1}{24} \\ & =\dfrac{(24+1) 5^{2 k}-1}{24} \\ & =\dfrac{24.5^{2 k}+5^{2 k}-1}{24} \\ & =\dfrac{24.5^{2 k}}{24}+\dfrac{5^{2 k}-1}{24} \\ & =5^{2 k}+\dfrac{5^{2 k}-1}{24}\end{align} Clearly $5^{2 k} \in \mathbb{Z} \quad \forall k \in \mathbf{N}$,
$\dfrac{5^{2 k}-1}{24} \in \mathbb{Z}$ by (a).
Thus the given statement is true for $n=k+1$. Hence by mathematical induction it is true for all $n \in \mathbf{N}$.
Question 12(ii)
Show by mathematical induction that $\dfrac{10^{n+1}-9 n-10}{81}$ is an integer.
Solution
1. For $n=1$ then \begin{align}\dfrac{10^{n+1}-9 n-10}{81}&=\dfrac{10^{i+1}-9.1-10}{81} \\ & =\dfrac{100-9-10}{81}=1 \in \mathbb{Z}\end{align} Thus it is true for $n=1$.
2. Let it be true for $n=k$, then \begin{align}\dfrac{10^{k+1}-9 k-10}{81} \in \mathbb{Z}\cdots (i)\end{align} 3. For $n=k+1$, then we have \begin{align}\dfrac{10^{k+1+1}-9(k+1)-10}{81}& =\dfrac{10^{k+1}(10)-9 k-9-10}{81} \\ & =\dfrac{10^{k+1}(9+1)-9 k-9-10}{81} \\ & =\dfrac{9.10^{k+1}+10^{k+1}-9 k-9-10}{81} \\ & =\dfrac{9.10^{k+1}-9+10^{k+1}-9 k-10}{81} \\ & =\dfrac{9(10^{k+1}-1)+9+10^{k+1}-9 k-10}{81}\\ & =\dfrac{9(10^k+1-1)}{81}+\dfrac{10^{k+1}-9 k-10}{81} \\ & =\dfrac{10^{k+1}-1}{9}+\dfrac{10^{k+1}-9 k-10}{81}\end{align} Now $\dfrac{10^{k+1}-1}{9} \in \mathbb{Z} \forall k \in \mathrm{N}$ and $\dfrac{10^{k+1}-9 k-10}{81} \in \mathbb{Z} \quad$ by (i),
hence it is true for $n=k+1$. Thus by mathematical induction it is true for all $n \in \mathbf{N}$.
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