a n − b n is always divisible by Options: A . &nbsp 2a-b B . &nbsp a+b C . &nbsp a-b D . &nbsp a-2b

an−bn is always divisible by

Discussion Forum : Principle Of Mathematical Induction

Question -

$a^{n} - b^{n}$ is always divisible by

Options:

A . 2a-b

B . a+b

C . a-b

D . a-2b

Answer: Option C
:
C

Substituting n=1, we get
$a^{1} - b^{1} = a - b$
Let $P (n) : a^{n} - b^{n} is divisible by a-b$
P(1) is true
Assume P(k) is true
$a^{k} - b^{k} is divisible by a-b a^{k} - b^{k} = m (a - b) N o w, a^{k + 1} - b^{k + 1} = a . (a^{k} - b^{k}) + (a - b) b^{k} = (a m + b^{k}) (a - b) \to divisible by a-b$
P(k+1) is true
Hence, P(n) is true.

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Discussion Forum : Principle Of Mathematical Induction

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