For all natural numbers n, 10 2 n − 1 + 1 is divisible by Options: A . &nbsp 13 B . &nbsp 11 C . &nbsp 9 D . &nbsp none of these

Answer: Option B : B Substituting n=1, we have 10 1 + 1 = 11 L e t P ( n ) : 10 2 n − 1 + 1 is divisible by 11 P(1) is true. Assume P(k) is true ⇒ 10 2 k − 1 + 1 = 11 m N o w , P ( k + 1 ) : 10 2 k + 1 + 1 = 100.10 2 k − 1 + 1 = 99.10 2 k − 1 + 10 2 k − 1 + 1 11. ( 9.10 2 k − 1 + m ) → divisible by 11 P(k+1) is also true. Hence, P(n) is true.

For all natural numbers n, 102n−1+1 is divisible by

Discussion Forum : Principle Of Mathematical Induction

Question -

For all natural numbers n, $10^{2 n - 1} + 1$ is divisible by

Options:

A . 13

B . 11

C . 9

D . none of these

Answer: Option B
:
B

Substituting n=1, we have
$10^{1} + 1 = 11 L e t P (n) : 10^{2 n - 1} + 1 is divisible by 11$
P(1) is true.
Assume P(k) is true
$\Rightarrow 10^{2 k - 1} + 1 = 11 m N o w, P (k + 1) : 10^{2 k + 1} + 1 = {100.10}^{2 k - 1} + 1 = {99.10}^{2 k - 1} + 10^{2 k - 1} + 1 11. ({9.10}^{2 k - 1} + m) \to divisible by 11$
P(k+1) is also true.
Hence, P(n) is true.

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Discussion Forum : Principle Of Mathematical Induction

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