Question

question no. 23.....!!!!

Answer 1

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23)
$= > p(n) \bf : {41}^{n} - {14}^{n} \\ \\ = > p(1) \bf : 41 - 14 = 27$
Which is a multiple of 27.

Hence p(1) is true.

Let p(k) is also true.
$= > p(k) : \bf \: {41}^{k} - {14}^{k} = 27m \\ \\ = > p(k + 1) \bf : {41}^{k + 1} - {14}^{k + 1} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 41 \times {41}^{k} - 14 \times {14}^{k} \\ \\ \: \: \: \: \: \: \: \: \: \: = 41(27m + {14}^{k } ) - 14 \times {14}^{k} \\ \\ \: \: \: \: \: \: \: \: =( 27 \times {41}) + 41 \times {14}^{k} - 14 \times {14}^{k} \\ \\ \: \: \: \: \: = (27 \times 41) + {14}^{k} (41 - 14) = (27 \times 41) +( 27 \times {14}^{k} ) \\ \\ \bf \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 27(41 + {14}^{k} )$

Which is a multiple of 27.

Hence, p(n) is always true for all natural numbers (N).

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