Question

show that 3 · 4^n + 51 is divisible by 3 and 9 for all positive integers n by PMI​

Answer 1

$\text{Let P(n) denote the statement}$

$P(n):\;3\,(4)^n+51\;\text{is divisible by 3}$

$\text{For n=1,}$

$P(1):\;3\,(4)^1+51=3(4)+51=63\;\text{whihch is divisible by 3}$

$\therefore\,P(1)\,\text{is true}$

$\text{Assume the result is true for n=k}$

$\text{That is,}\;3\,(4)^k+51\;\text{is divisible by 3 is true}$

$\text{Then,}\,3\,(4)^k+51=3\,c\;\;\text{where c is an integer}$

$\text{To prove:}\;P(k+1)\,\text{is true}$

$P(k+1):$

$3\,(4)^{k+1}+51$

$=3\,(4)^k4+51$

$=(3c-51)\,4+51$

$=12c-204+51$

$=12c-153$

$=3(4c-51)\;\text{which is divisible by 3}$

$\therefore\,P(k+1)\;\text{is true}$

$\textbf{Hence by principal of mathematical induction P(n) is true}$

$\textbf{for all natrural numbers}$

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Answer 2

Given:

3 · 4ⁿ + 51

To prove:

3 · 4ⁿ + 51 is divisible by 3 and 9 for all positive integers n

Proof:

we have to show the above statement by the PMI

For this we have

3 · 4ⁿ + 51

Case 1: Put the value of n = 1

• 3 · 4ⁿ + 51
• 3 · 4¹ + 51
• 12 + 51
• 63

as we know that 63 is the multiple of 3 and 9, so it is true for n=1.

Case 2: Let the condition is true for n = a

• 3 · 4ᵃ + 51
• 3 · 4ᵃ + 3.17
• 3 (4ᵃ + 17)

So it is divisible by 3 and 9

Case 3: Now prove the condition is true for n = a+1

• $3.4^{(a+1)}+51$
• $3.4^{a}.4+51$
• $3.4^{a}.4+3.17$
• $3.(4^{a}.4+17)$
• $3.(4^{a}+17+ 3.4^{a})}$

(4ᵃ + 17) is divisible by 3 so it can be represented as 3k

• $3.(3K+ 3.4^{a})}$
• $9.(K+ 4^{a})}$

Hence it is also divisible by 3 and 9

Hence proved.