If $n > 1$ , then $(1 + x)^{n} - n x - 1$ is divisible by:

x^{5}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

x^{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $x^{2}$
$(1 + x)^{n} = 1 +^{n} C_{1} x +^{n} C_{2} x^{2} + . . . . . . . +^{n} C_{n} x^{n}$
$(1 + x)^{n} - n x - 1$
$= (1 +^{n} C_{1} x + . . . . . . . . +^{n} C_{n} x^{n}) - n x - 1$
$=^{n} C_{2} x^{2} + . . . . . . +^{n} C_{n} x^{n}$
This is divisible by $x^{2}$

Suggest Corrections

Similar questions

(1 + x)^{n} - n x - 1

(w h e r e n ϵ N)

(x + 1)^{n} - x^{n} - 1

x^{3} + x^{2} + x,

n > 1

x \neq 0

{(1 + x)}^{n} - n x - 1

$x^{5} + x^{4} + x^{3} + x^{2} + x + 1$ by $x^{3} + 1$

{(1 + x)}^{n} - n x - 1

x^{2}, n \in N

(x^{n} - y^{n})

(x - y) w h e r e x \neq y a n d n \in N .

Join BYJU'S Learning Program