Question

please solve the above question(only for Advanced Maths)​

Answer 1

Answer:

Inductive Step:  

Assuming that this is true for n = k,  Then it should be true for (k + 1) also,

3^(4(k+1) + 2) + 5^(2(k+1) + 1)

= 3^(4k+6) + 5^(2k+3)

= 3^4 * 3^(4k+2) + 5^2 * 5^(2k+1)

= 81 * 3^(4k+2) + 25 * 5^(2k+1)

= (25 + 56) * 3^(4k+2) + 25 * 5^(2k+1)

= 25 [3^(4k+2) + 5^(2k+1)] + 56 * 3^(4k+2).

By inductive hypothesis, 3^(4k+2) + 5^(2k+1) is divisible by 14.

Clearly, 56 * 3^(4k+2) = 14 * 4 * 3^(4k+2) is divisible by 14.

Therefore, 3^(4(k+1) + 2) + 5^(2(k+1) + 1)

= 25 [3^(4k+2) + 5^(2k+1)] + 56 * 3^(4k+2) is divisible by 14, completing the inductive step.

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Answer 2

GIVEN

$\bullet \tt\: \: f(n) = {3}^{4n + 2} + 5 {}^{2n + 1}$

TO PROVE

• f(n) is divisible by 14

PROOF

Taking n = 1 we have ,

$\tt \implies f(1) = {3}^{4 \times 1 + 2} + {5}^{2 \times 1 + 1} \\ \\ \tt \implies f(1) = {3}^{6} + 5 {}^{3} \\ \\ \tt \implies f(1) = 729 + 125 \\ \\ \tt \implies f(1) = 854 \\ \\ \tt \implies f(1) = 61 \times 14 \\ \\ \tt Hence \: f(n) \: is \: divisible \: by \: 14 \\ \tt for \: \: n = 1$

$\tt Now \: let \: us \: take \: \\ \tt f(k) = {3}^{4k + 2} + {5}^{2n + 1} \: \: is \: divisible \\ \tt by \: 14 \\ \tt Thus \: \: f(k) = 14p \: \: \{ where \: p \: be \: any \: constant\}$

$\tt {Again} \: \: f(k + 1) \: \: will \: be \: \\ \\ \tt = {3}^{4(k + 1) + 2} + {5}^{2(k + 1) + 1} \\ \\ \tt = {3}^{4k + 2 + 4} + {5}^{2k + 1 + 2} \\ \\ = \tt {3}^{4k + 2} \times {3}^{4} + {5}^{2k + 1} \times {5}^{2} \\ \\ \tt= 81 \times {3}^{4k + 2} + 25 \times 5 {}^{2k + 1} \\ \\ \tt = (56 + 25) {3}^{4k + 2} + 25 \times {5}^{2k + 1} \\ \\ \tt = 56 \times {3}^{4k + 2} + 25 \times {3}^{4k + 2} + 25 \times {5}^{2k + 1} \\ \\ \tt = 14 \times 4 \times {3}^{4k + 2} + 25(3 {}^{4k + 2} + {5}^{2k + 1} ) \\ \\ \tt = 14 \times 4 \times {3}^{4k + 2} + 25 \times 14p$

$\tt Thus \: f(k + 1) \: is \: also \: divided \: \\ \tt by \: 14$

$\tt Hence \: by \: the \: method \: of \: induction \\ \tt f(n) = {3}^{4n + 2} + {5}^{2n + 1} \: is \: divided \\ \tt by \: 14$