Class 11: Mathematical Induction

$\displaystyle \textbf{Question 1: }\text{If }P(n)\text{ is the statement that }n(n+1)\text{ is even, then what is }P(3)\text{?}$
$\displaystyle \text{Answer:}$
$\displaystyle P(n):n(n+1)\text{ is even.}$
$\displaystyle P(3):3(3+1)\text{ is even.}$
$\displaystyle 3(3+1)=3\times4=12$
$\displaystyle \text{Since }12\text{ is even, }P(3)\text{ is true.}$
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$\displaystyle \textbf{Question 2: }\text{If }P(n)\text{ is the statement that }n^3+n\text{ is divisible by }3,\text{ prove that}$
$\displaystyle P(3)\text{ is true but }P(4)\text{ is not true.}$
$\displaystyle \text{Answer:}$
$\displaystyle P(n):n^3+n\text{ is divisible by }3.$
$\displaystyle P(3):3^3+3=27+3=30$
$\displaystyle \text{Since }30\text{ is divisible by }3,\ P(3)\text{ is true.}$
$\displaystyle P(4):4^3+4=64+4=68$
$\displaystyle \text{Since }68\text{ is not divisible by }3,\ P(4)\text{ is not true.}$
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$\displaystyle \textbf{Question 3: }\text{If }P(n)\text{ is the statement }2^n\geq3n\text{ and }P(r)\text{ is true, prove that}$
$\displaystyle P(r+1)\text{ is true.}$
$\displaystyle \text{Answer:}$
$\displaystyle P(n):2^n\geq3n$
$\displaystyle \text{Given that }P(r)\text{ is true.}$
$\displaystyle \therefore 2^r\geq3r$
$\displaystyle \text{Multiplying both sides by }2,$
$\displaystyle 2^{r+1}\geq6r$
$\displaystyle \text{Since }r\geq1,\text{ we have }3r\geq3.$
$\displaystyle \therefore 6r=3r+3r\geq3r+3=3(r+1)$
$\displaystyle \therefore 2^{r+1}\geq3(r+1)$
$\displaystyle \therefore P(r+1)\text{ is true.}$
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$\displaystyle \textbf{Question 4: }\text{If }P(n)\text{ is the statement that }n^2+n\text{ is even and }P(r)\text{ is true,}$
$\displaystyle \text{prove that }P(r+1)\text{ is true.}$
$\displaystyle \text{Answer:}$
$\displaystyle P(n):n^2+n\text{ is even.}$
$\displaystyle \text{Given that }P(r)\text{ is true.}$
$\displaystyle \therefore r^2+r\text{ is even.}$
$\displaystyle \therefore r^2+r=2\lambda,\text{ where }\lambda\in N$
$\displaystyle P(r+1):(r+1)^2+(r+1)$
$\displaystyle =r^2+2r+1+r+1$
$\displaystyle =(r^2+r)+2(r+1)$
$\displaystyle =2\lambda+2(r+1)$
$\displaystyle =2(\lambda+r+1)$
$\displaystyle \text{Let }\mu=\lambda+r+1.$
$\displaystyle \therefore (r+1)^2+(r+1)=2\mu$
$\displaystyle \therefore (r+1)^2+(r+1)\text{ is even.}$
$\displaystyle \therefore P(r+1)\text{ is true.}$
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$\displaystyle \textbf{Question 5: }\text{Give an example of a statement }P(n)\text{ which is true for all } \\ n\in N.$
$\displaystyle \text{Answer:}$
$\displaystyle P(n):1+2+3+\ldots+n=\frac{n(n+1)}{2}$
$\displaystyle \text{This statement is true for all }n\in N.$
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$\displaystyle \textbf{Question 6: }\text{If }P(n)\text{ is the statement that }n^2-n+41\text{ is prime, prove that}$
$\displaystyle P(1),\ P(2)\text{ and }P(3)\text{ are true. Also prove that }P(41)\text{ is not true.}$
$\displaystyle \text{Answer:}$
$\displaystyle P(n):n^2-n+41\text{ is prime.}$
$\displaystyle P(1):1^2-1+41=41$
$\displaystyle \text{Since }41\text{ is prime, }P(1)\text{ is true.}$
$\displaystyle P(2):2^2-2+41=4-2+41=43$
$\displaystyle \text{Since }43\text{ is prime, }P(2)\text{ is true.}$
$\displaystyle P(3):3^2-3+41=9-3+41=47$
$\displaystyle \text{Since }47\text{ is prime, }P(3)\text{ is true.}$
$\displaystyle P(41):41^2-41+41=41^2=1681$
$\displaystyle \text{Since }1681=41\times41,\text{ it is not prime.}$
$\displaystyle \therefore P(41)\text{ is not true.}$
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$\displaystyle \textbf{Question 7: }\text{Give an example of a statement }P(n)\text{ which is true for all }$
$\displaystyle n\geq4, \ \text{but }P(1),\ P(2)\text{ and }P(3)\text{ are not true. Justify your answer.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }P(n):3n<n!$
$\displaystyle P(1):3\times1<1!\Rightarrow3<1,\text{ which is false.}$
$\displaystyle \therefore P(1)\text{ is not true.}$
$\displaystyle P(2):3\times2<2!\Rightarrow6<2,\text{ which is false.}$
$\displaystyle \therefore P(2)\text{ is not true.}$
$\displaystyle P(3):3\times3<3!\Rightarrow9<6,\text{ which is false.}$
$\displaystyle \therefore P(3)\text{ is not true.}$
$\displaystyle P(4):3\times4<4!\Rightarrow12<24,\text{ which is true.}$
$\displaystyle P(5):3\times5<5!\Rightarrow15<120,\text{ which is true.}$
$\displaystyle \text{Since }n!\text{ grows faster than }3n,\text{ the statement is true for all }n\geq4.$
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