Find the coefficient of x^{49} in the expansi | vedclass

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

Question

(B) The given expression is $P(x) = (2x + 1)(2x + 3)(2x + 5) \dots (2x + 99)$.Factor out $2$ from each of the $50$ terms: $P(x) = 2^{50} \left(x + \fr

Vedclass · Accepted Answer

$2^{49} 	imes 2500$

Vedclass · Answer

(B) The given expression is $P(x) = (2x + 1)(2x + 3)(2x + 5) \dots (2x + 99)$.Factor out $2$ from each of the $50$ terms: $P(x) = 2^{50} \left(x + \frac{1}{2}ight) \left(x + \frac{3}{2}ight) \dots \left(x + \frac{99}{2}ight)$.Let $a_k = \frac{2k-1}{2}$ for $k = 1, 2, \dots, 50$. Then $P(x) = 2^{50} \prod_{k=1}^{50} (x + a_k)$.The coefficient of $x^{49}$ in the expansion of $\prod_{k=1}^{50} (x + a_k)$ is the sum of the roots taken $49$ at a time,which is equivalent to the sum of all roots multiplied by $(-1)$ if we consider the Vieta's formula for a polynomial of degree $50$. Specifically,the coefficient of $x^{n-1}$ in $\prod (x+a_k)$ is $\sum a_k$.Thus,the coefficient of $x^{49}$ is $2^{50} 	imes \sum_{k=1}^{50} \frac{2k-1}{2} = 2^{49} 	imes \sum_{k=1}^{50} (2k-1)$.The sum of the first $50$ odd numbers is $50^2 = 2500$.Therefore,the coefficient is $2^{49} 	imes 2500$.

Find the coefficient of $x^{49}$ in the expansion of $(2x + 1)(2x + 3)(2x + 5) \dots (2x + 99)$.

Similar Questions

If the number of terms in the binomial expansion of $(2x + 3)^{3n}$ is $22$,then the value of $n$ is:

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

If $A$ denotes the sum of all the coefficients in the expansion of $(1-3x+10x^2)^n$ and $B$ denotes the sum of all the coefficients in the expansion of $(1+x^2)^n$,then:

The last two digits of the number $3^{400}$ are:

Find the value of $(a^{2}+\sqrt{a^{2}-1})^{4}+(a^{2}-\sqrt{a^{2}-1})^{4}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module