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Euclid's Division Lemma

72 n+23 n-3· ...

Question

7²ⁿ + 2³ⁿ⁻³. 3ⁿ⁻¹ is divisible by 25 for all n ∈ N.

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Solution

Let P(n) be the given statement.
Now,
$P (n) : 7^{2 n} + 2^{3 n - 3} . 3^{n - 1} is divisible by 25 . Step 1 : P (1) : 7^{2} + 2^{3 - 3} . 3^{1 - 1} = 49 + 1 = 50 I t is divisible by 25 . Thus, P (1) is true Step 2 : Let P (m) be true . N o w, 7^{2 m} + 2^{3 m - 3} . 3^{m - 1} is divisible by 25 . Suppose : 7^{2 m} + 2^{3 m - 3} . 3^{m - 1} = 25 λ . . . (1) We have to show that P (m + 1) is true whenever P (m) is true . Now, P (m + 1) = 7^{2 m + 2} + 2^{3 m} . 3^{m} = 7^{2 m + 2} + 7^{2} . 2^{3 m - 3} . 3^{m - 1} - 7^{2} . 2^{3 m - 3} . 3^{m - 1} + 2^{3 m} . 3^{m} = 7^{2} (7^{2 m} + 2^{3 m - 3} . 3^{m - 1}) + 2^{3 m} . 3^{m} (1 - \frac{49}{24}) = 7^{2} \times 25 λ - 2^{3 m} . 3^{m} \times \frac{25}{2^{3} . 3^{1}} [Using (1)] = 25 (49 λ - 2^{3 m - 3} . 3^{m - 1}) I t is divisible by 25 . Thus, P (m + 1) is true . B y t h e p rinciple of m athematical i nduction, P (n) is true for all n \in N .$

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Similar questions

$7^{2 n} + 2^{3 n - 3} {.3}^{n - 1}$ is divisible by 25 for all $n ϵ$ N.

7^{2 n} + 2^{3 n - 3} {.3}^{n - 1}

is divisible by

25

for any natural number

n \geq 1

. Prove that by mathematical

n \in N

7^{2 n} + 2^{3 n - 3} . 3^{n - 1}

is always divisible by

7^{2 n} + 3^{n - 1} \cdot 2^{3 n - 3}

is divisible by

If n ∈ N, then $7^{2 n}$ + $2^{3 n - 3}$ . $3^{n - 1}$ is always divisible by

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