Question

For all ne N. prove that 41^n- 14^n is a multiple of 27.​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

$\bold {\huge {\underline {\underline {\sf { ANSWER}}}}}$

$\sf {41}^{n}$ - $\sf {14}^{n}$ is a multiple of 27

$\sf {\bold {\underline {Step\: I : Put \:n = 1 }}}$

$\sf {41}^{1}$ - $\sf {14}^{1}$

= 41 - 14

= 27

$\therefore$ P ( 1 ) is true .

$\sf{\bold {\underline {Step \:II : Assume\: P ( k )\: is \:true}}}$

i.e. $\sf {41}^{k}$ - $\sf {14}^{k} = \sf 27m$

$\therefore$ $\sf {41}^{k}$ = $\sf 27m$ + $\sf {14}^{k}$ ... ( 1 )

$\sf{\bold {\underline {Step\:III : Prove \: P (k+1) \: is \: true}}}$

$\therefore$$\sf {41}^{k+1}$ - $\sf {14}^{k+1}$ = 27n

$\therefore$$\sf{41}^{k}.41$ - $\sf{14}^{k}.14$

$\therefore$[tex]\sf