Question

Please answer it and explain

Answer 1

1. Postive integer...

is the correct answer...

Answer 2

Answer:

Option(3)

Step-by-step explanation:

Let the given statement be P(n). Then,

P(n) = (xⁿ + aⁿ) is divisible by x + a.

When n = 1:

LHS : x¹ + a¹ = x + a

RHS : x + a

Thus, the given statement is true for n = 1, i.e P(1) is true.

Let P(2k + 1) is true. Then,

P(2k + 1) : x^(2k + 1) + a^(2k + 1) is divisible by (x + a).

Now,

$\Longrightarrow \frac{x^{2k+1} + a^{2k + 1}}{x + a}$

$\Longrightarrow\frac{x^{2k} * x + a^{2k} * a}{x + a}$

$\Longrightarrow\frac{xx^{2k}+aa^{2k}+x^{2k}a-x^{2k}a + a^{2k}x-a^{2k}x}{x+y}$

$\Longrightarrow\frac{x^{2k}(x+a)+a^{2k}(x+a)-x^{2k}a- a^{2k}a}{x+y}$

$\Longrightarrow\frac{x^{2k}(x+y)+y^{2k}(x+y)-xy(x^{2k-1}+y^{2k-1})}{x+y}$

$\Longrightarrow P(2k + 1) : x^{2k-1} + a^{2k-1} \;is \;divisible \; by \;x + a$

$\Longrightarrow P(2k + 1) \; is \; true, whenever \; P(k) \; is \; true$

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence,by the mathematical induction, P(n) is true for all odd positive integer.

Hope it helps!