(1 + x)^n - nx - 1 is divisible by (where n i | vedclass

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

Question

(B) Using the binomial expansion: $(1 + x)^n = 1 + nx + \frac{n(n - 1)}{2!}x^2 + \frac{n(n - 1)(n - 2)}{3!}x^3 + \dots$ Subtracting $nx + 1$ from both

Vedclass · Accepted Answer

$x^2$

Vedclass · Answer

(B) Using the binomial expansion: $(1 + x)^n = 1 + nx + \frac{n(n - 1)}{2!}x^2 + \frac{n(n - 1)(n - 2)}{3!}x^3 + \dots$ Subtracting $nx + 1$ from both sides: $(1 + x)^n - nx - 1 = \frac{n(n - 1)}{2!}x^2 + \frac{n(n - 1)(n - 2)}{3!}x^3 + \dots$ Factoring out $x^2$: $(1 + x)^n - nx - 1 = x^2 \left[ \frac{n(n - 1)}{2!} + \frac{n(n - 1)(n - 2)}{3!}x + \dots \right]$ Since the expression can be written as $x^2$ multiplied by a polynomial in $x$,it is clearly divisible by $x^2$.

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

Similar Questions

The sum of the coefficients in the expansion of $(1+x+x^2)^n$ is

Using the binomial theorem, the value of $(0.999)^3$ correct to $3$ decimal places is:

The total number of terms in the expansion of $\left[ (1 + x)^{100} + (1 + x^2)^{100}(1 + x^3)^{100} \right]$ is

Expand the expression $\left(x+\frac{1}{x}\right)^{6}$.

If the first three terms in the binomial expansion of $(1+bx)^n$ in ascending powers of $x$ are $1, 6x$ and $6x^2$ respectively,then $b+n=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module