Question

question no. 22 and 23.

Answer 1

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Answer 2

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$= > \bf \: p(n) : {3}^{2n + 2} - 8n - 9 \\ \\ \bf \: = > p(1) : {3}^{2 + 2} - 8 - 9 = 81 - 17 = 64$
Which is divisible by 8.

Hence p(1) is true.

Let p(k) is also true.

$= > \bf \:p(k) : {3}^{2k + 2} - 8k - 9 \\ \\ = > \bf \: p(k + 1) : {3}^{2(k + 1) + 2} - 8(k + 1) - 9 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = {3}^{2} \times {3}^{2k + 2} - 72k + 64k - 81 + 72 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 9 ( {3}^{2k + 2} - 8k - 9) + 8(8k + 9) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 9 \times 8n + 8(8k + 9) = 8(9n + 8k + 9)$
Which is a multiple of 8

Hence p(m) is true for all natural number (N)
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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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