3 Marks Question

Question 13 Marks

Give an example of a statement P(n) which is true for all $\text{n}\geq4$ but P(1), P(2) and P(3) are not true. Justify your answer.

Answer

Consider the statement P(n): 3n < n!
For n = 1, 3 × 1 < 1!, which is not true.
For n = 2, 3 × 2 < 2!, which is not true.
For n = 3, 3 × 3 < 3!, which is not true.
For n = 4, 3 × 4 < 4!, which is true.
For n = 5, 3 × 5 < 5!, which is true.

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Question 23 Marks

Prove the following statement by principle of mathematical induction:
For any natural number $n, 7^n - 2^n$ is divisible by $5.$

Answer

Let $\text{P}(n): 7^n - 2^n$
Step $1: \text{P}(1): 7^1 - 2^1 = 5$ is divisible by $5.$
So$, \text{P}(1)$ is true.
Step $2:$ Assume $\text{P}(k)$ is true some $\text{k }\epsilon\text{ N}\Rightarrow\text{P(k): }7^{\text{k}}-2^{\text{k}}=5\lambda,\lambda\ \epsilon\text{ N}\ ...(\text{i})$
Step $3:$ Now we have to prove $\text{P}(k + 1): 7k+1 - 2k+1$ is divisible by $5.$
$\text{P(k + 1): }7^{\text{k}+1}-2^{\text{k}+1}$
$=7^{\text{k}+1}+7^{\text{k}}\cdot2-7^{\text{k}}\cdot2-2^{\text{k}+1}$
$=\big(7^{\text{k}+1}-7^{\text{k}}\cdot2\big)+\big(7^{\text{k}}\cdot2-2^{\text{k}+1}\big)$
$=7^{\text{k}}(7-2)+2\cdot(7^{\text{k}}-2^{\text{k}})$
$=5\cdot7^{\text{k}}+2\cdot5\lambda ($Using equation $(i))$
$=5\big(7^{\text{k}}+2\lambda\big)$ is divisible by $5.$
$\Rightarrow\text{P}(\text{k}+1)$ is true.
Hence$, \text{P}(k + 1)$ is true whenever $\text{P}(k)$ is true.
Therefore by the principle of mathematical induction we have $\text{P}(n)$ is true for all n.

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