The expression (1 + x)^n - nx - 1 is divisibl | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Using the Binomial Theorem,we expand $(1 + x)^n$ as follows:$(1 + x)^n = 1 + nx + \frac{n(n - 1)}{2!}x^2 + \frac{n(n - 1)(n - 2)}{3!}x^3 + \dots$N

Vedclass · Accepted Answer

by $x^2$

Vedclass · Answer

(B) Using the Binomial Theorem,we expand $(1 + x)^n$ as follows:$(1 + x)^n = 1 + nx + \frac{n(n - 1)}{2!}x^2 + \frac{n(n - 1)(n - 2)}{3!}x^3 + \dots$Now,subtract $nx + 1$ from the expansion:$(1 + x)^n - nx - 1 = (1 + nx + \frac{n(n - 1)}{2}x^2 + \frac{n(n - 1)(n - 2)}{6}x^3 + \dots) - nx - 1$Simplifying the expression:$(1 + x)^n - nx - 1 = x^2 \left[ \frac{n(n - 1)}{2} + \frac{n(n - 1)(n - 2)}{6}x + \dots \right]$Since the expression is a multiple of $x^2$,it is divisible by $x^2$.

The expression $(1 + x)^n - nx - 1$ is divisible by (where $n \in N$ and $n > 1$):

Similar Questions

If $2^{4n+3} + 3^{3n+1}$ is divisible by $P$ for all natural numbers $n$,then $P$ is

The number $(101)^{100}-1$ is divisible by

The remainder obtained when $5^{99}$ is divided by $13$ is

The remainder left when $8^{2n} - 62^{2n+1}$ is divided by $9$ is:

If $P(n):$ " $2^{2n}-1$ is divisible by $k$ for all $n \in N$ " is true,then the value of $k$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module