For n ∈ N, xn+1+(x+1)2n-1 is divisible by

Question

For n ∈ N, xn+1+(x+1)2n-1 is divisible by (A) x (B) x + 1 (C) x2 + x + 1 (D) x2 - x + 1

Tardigrade Admin · Accepted Answer

For n = 1, we have xn+1 + (x + 1)2n-1 = x2 + (x + 1) = x2 + x + 1, which is divisible by x2 + x + 1 For n = 2, we have xn+1 + (x + 1)2n-1 = x3 + (x + 1)3 = (2x + 1) (x2 + x + 1), which is divisible by x2 + x + 1.

Q. For $n \in N, x^{n + 1} + (x + 1)^{2 n - 1}$ is divisible by

$x$

$x + 1$

$x^{2} + x + 1$

$x^{2} - x + 1$

For $n = 1,$ we have
$x^{n + 1} + (x + 1)^{2 n - 1} = x^{2} + (x + 1) = x^{2} + x + 1,$
which is divisible by $x^{2} + x + 1$
For $n = 2$ , we have
$x^{n + 1} + (x + 1)^{2 n - 1} = x^{3} + (x + 1)^{3} = (2 x + 1) (x^{2} + x + 1),$ which is divisible by $x^{2} + x + 1.$

Q. For n∈N,xn+1+(x+1)2n−1 is divisible by

x

x+1

x2+x+1

x2−x+1

For n=1, we have xn+1+(x+1)2n−1=x2+(x+1)=x2+x+1, which is divisible by x2+x+1 For n=2, we have xn+1+(x+1)2n−1=x3+(x+1)3=(2x+1)(x2+x+1), which is divisible by x2+x+1.

Q. For $n \in N, x^{n + 1} + (x + 1)^{2 n - 1}$ is divisible by

$x$

$x + 1$

$x^{2} + x + 1$

$x^{2} - x + 1$

For $n = 1,$ we have
$x^{n + 1} + (x + 1)^{2 n - 1} = x^{2} + (x + 1) = x^{2} + x + 1,$
which is divisible by $x^{2} + x + 1$
For $n = 2$ , we have
$x^{n + 1} + (x + 1)^{2 n - 1} = x^{3} + (x + 1)^{3} = (2 x + 1) (x^{2} + x + 1),$ which is divisible by $x^{2} + x + 1.$