For all n - N - 3 - 52 n-1-23 n-1 is divisible byA- 17B- 23C- 25D- 19

The correct option is B 23 (b) and (c) Given that, 3 × 5^2n + 1 + 2^3n + 1 For n = 1 =3 × 5^2(1) + 1 + 2^3(1) + 1 = 3 × 5^3 + 2^4 = 3 × 125 + 16 = 375 + 16 = 3 ...

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Standard XII

Mathematics

Proof by mathematical induction

For all n ∈ N...

Question

For all $n ϵ N, 3 \times 5^{2 n + 1} + 2^{3 n + 1}$ is divisible by

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Solution

The correct options are
B
17

C
23

(b) and (c)

Given that, $3 \times 5^{2 n + 1} + 2^{3 n + 1}$

For n = 1

$= 3 \times 5^{2 (1) + 1} + 2^{3 (1) + 1}$

$= 3 \times 5^{3} + 2^{4}$

$= 3 \times 125 + 16 = 375 + 16 = 391$

Now, 391 = 17 $\times$ 23

which is divisible by both 17 and 23.

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For all nϵN, 3×52n+1+23n+1 is divisible by

The correct options are B 17 C 23 (b) and (c) Given that, 3×52n+1+23n+1 For n = 1 =3×52(1)+1+23(1)+1 =3×53+24 =3×125+16=375+16=391 Now, 391 = 17 × 23 which is divisible by both 17 and 23.

For all $n ϵ N, 3 \times 5^{2 n + 1} + 2^{3 n + 1}$ is divisible by

The correct options are
B
17

C
23

(b) and (c)

Given that, $3 \times 5^{2 n + 1} + 2^{3 n + 1}$

For n = 1

$= 3 \times 5^{2 (1) + 1} + 2^{3 (1) + 1}$

$= 3 \times 5^{3} + 2^{4}$

$= 3 \times 125 + 16 = 375 + 16 = 391$

Now, 391 = 17 $\times$ 23

which is divisible by both 17 and 23.