Given (1 - 2x + 5x^2 - 10x^3) (1 + x)^n = 1 + | vedclass

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

Question

(A) We are given the expansion $(1 - 2x + 5x^2 - 10x^3)(1 + x)^n = 1 + a_1x + a_2x^2 + \dots$Using the binomial expansion $(1+x)^n = 1 + nx + \frac{n(

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) We are given the expansion $(1 - 2x + 5x^2 - 10x^3)(1 + x)^n = 1 + a_1x + a_2x^2 + \dots$Using the binomial expansion $(1+x)^n = 1 + nx + \frac{n(n-1)}{2}x^2 + \dots$,we have:$(1 - 2x + 5x^2 - 10x^3)(1 + nx + \frac{n(n-1)}{2}x^2 + \dots) = 1 + a_1x + a_2x^2 + \dots$Comparing the coefficients of $x$:$a_1 = n - 2$Comparing the coefficients of $x^2$:$a_2 = \frac{n(n-1)}{2} - 2n + 5$Given the condition $a_1^2 = 2a_2$:$(n - 2)^2 = 2 \left( \frac{n(n-1)}{2} - 2n + 5 \right)$$n^2 - 4n + 4 = n(n-1) - 4n + 10$$n^2 - 4n + 4 = n^2 - n - 4n + 10$$n^2 - 4n + 4 = n^2 - 5n + 10$$n = 6$

Given $(1 - 2x + 5x^2 - 10x^3) (1 + x)^n = 1 + a_1x + a_2x^2 + \dots$ and that $a_1^2 = 2a_2$,then the value of $n$ is:

Similar Questions

If $f(x) = x^n$,then the value of $f(1) - \frac{f'(1)}{1!} + \frac{f''(1)}{2!} - \frac{f'''(1)}{3!} + ...... + \frac{(-1)^n f^n(1)}{n!}$ is

Find the coefficient of $x^5$ in $(1+x+x^2)^8$.

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

Which is larger: $(1.01)^{1000000}$ or $10,000$?

If $(1+x+x^2+x^3)^5 = \sum_{k=0}^{15} a_k x^k$,then $\sum_{k=0}^7 a_{2k}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module