3 Marks Question

Question 13 Marks

Prove the following by using the principle of mathematical induction for all n ∈ N:
$\Big(1+\frac{1}{1}\Big)\Big(1+\frac{1}{2}\Big)\Big(1+\frac{1}{3}\Big)...\Big(1+\frac{1}{\text{n}}\Big)=(\text{n+1}).$

Answer

Let the given statement be P(n), i.e.,
$\text{P(n)}:\Big(1+\frac{1}{1}\Big)\Big(1+\frac{1}{2}\Big)\Big(1+\frac{1}{3}\Big)...\Big(1+\frac{1}{\text{n}}\Big)=(\text{n+1})$
For n = 1, we have
$\text{P}(1):\Big(1+\frac{1}{1}\Big)=2=(1+1),$
which is true.
Let P(k) be true for some positive integer k, i.e.,
$\text{P(k)}:\Big(1+\frac{1}{1}\Big)\Big(1+\frac{1}{2}\Big)\Big(1+\frac{1}{3}\Big)...\Big(1+\frac{1}{\text{k}}\Big)=(\text{k+1})\ ....(1)$
We shall now prove that P(k + 1) is true.
Consider,
$\Big[\Big(1+\frac{1}{1}\Big)\Big(1+\frac{1}{2}\Big)\Big(1+\frac{1}{3}\Big)...\Big(1+\frac{1}{\text{k}}\Big)\Big]\Big(1+\frac{1}{\text{k}+1}\Big)$
$=(\text{k}+1)\Big(1+\frac{1}{\text{k}+1}\Big)\ \ [\text{Using (1)}]$
$=(\text{k}+1)\Big(\frac{(\text{k+1})+1}{(\text{k}+1)}\Big)$
$=(\text{k}+1)+1$
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

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Question 23 Marks

If P(n) is the statement "$n^2 + n$ is even", and if $P(r)$ is true, then $P(r + 1)$ is true.

Answer

$P(n): n^2 + n$ is even
Given, $P(r)$ is true
$\Rightarrow r^2 + r$ is even
$\Rightarrow \text{r}^2 + \text{r} = 2\lambda \ ...(1)$
Now,
$(r + 1)^2 + (r + 1)$
$= r^2 + 1 + 2r + r + 1$
$= (r + 1)^2 + 2r + 2$
$=2\lambda + 2\text{r} + 2$ [Using equation (1)]
$=2(\lambda + \text{r} + 1)$
$=2\lambda$
$\Rightarrow (r + 1)^2 + (r + 1) is even$
$\Rightarrow P(r + 1)$ is true

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Question 33 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 43 Marks

If $P(n)$ is the statement " $n^2-n+41$ is prime", prove that $P(1), P(2)$ and $P(3)$ are true. Prove also that $P(41)$ is not true.

Answer

$P(n): n^2 - n + 41$ is prime
$P(1): 1 - 1 + 41$ is prime
$\Rightarrow P(1): 41$ is prime
$\therefore$ P(1) is true.
$P(2): 2^2 - 2 + 41$ is prime
$\Rightarrow P(2): 43$ is prime
$\therefore P(2)$ is true.
$P(3): 3^2 - 3 + 41 is prime$
$\Rightarrow P(3): 47$ is prime
$\therefore$ P(3) is true.
$P(41): (41)^2 - 41 + 41$ is prime
$P(41): (41)^2 is prime$
$\Rightarrow P(41)$ is not true.

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Question 53 Marks

If P(n) is the statement $"2^n \geq 3n"$ and if $P(r)$ is true, prove that $P(r + 1)$ is true.

Answer

$P(n): 2^n \geq 3n$
Given that P(r) is true
$\Rightarrow 2^r \geq 3r$
Multiplying both sides by $2,$
$2.2^r \geq 2.3r$
$2^{r+1} \geq 6r$
$2^{r+1} \geq 3r + 3r$
$2^{r+1} \geq 3 + 3r,$ [Since $3r \geq 3 \Rightarrow 3r + 3r \geq 3 + 3r]$
$2^{r+1} \geq 3r(r + 1)$
$\Rightarrow P(r + 1)$ is true.

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Question 63 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 $P(n): 7^n - 2^n$
Step 1: $P(1): 7^1 - 2^1 = 5$ is divisible by $5.$
So, P(1) is true.
Step 2: Assume P(k) is true some $\text{k }\in\text{ N}\Rightarrow\text{P(k): }7^{\text{k}}-2^{\text{k}}=5\lambda,\lambda\ \in\text{ N}\ ...(\text{i})$
Step 3: Now we have to prove $P(k + 1): 7^{k+1} - 2^{k+1}$ 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, P(k + 1) is true whenever P(k) is true.
Therefore by the principle of mathematical induction we have P(n) is true for all n.

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