The coefficient of x^{15} in the product (1-x | vedclass

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

Question

(C) Let $P(x) = \prod_{k=0}^{15} (1-2^k x)$.We want the coefficient of $x^{15}$ in this product.This is a known identity related to the Gaussian binom

Vedclass · Accepted Answer

$2^{105}-2^{121}$

Vedclass · Answer

(C) Let $P(x) = \prod_{k=0}^{15} (1-2^k x)$.We want the coefficient of $x^{15}$ in this product.This is a known identity related to the Gaussian binomial coefficient or $q$-binomial theorem,where the coefficient of $x^n$ in $\prod_{k=0}^{n-1} (1-q^k x)$ is $(-1)^n q^{n(n-1)/2}$.Here,$n=15$ and $q=2$.The product is $\prod_{k=0}^{15} (1-2^k x) = (1-x)(1-2x)(1-4x)\ldots(1-2^{15}x)$.The coefficient of $x^{15}$ is $(-1)^{15} 	imes 2^{0+1+2+\ldots+14} = -1 	imes 2^{\frac{14 	imes 15}{2}} = -2^{105}$.However,checking the options provided,there seems to be a discrepancy in the standard form. Given the structure of the product,the coefficient of $x^{15}$ is $(-1)^{15} \sum_{0 \le i_1 Since we are picking $15$ terms out of $16$ available indices $(0, 1, \ldots, 15)$,the only way to pick $15$ is to exclude exactly one index $j \in \{0, 1, \ldots, 15\}$.The coefficient is $(-1)^{15} \sum_{j=0}^{15} \frac{\prod_{k=0}^{15} 2^k}{2^j} = -\frac{2^{\frac{15 	imes 16}{2}}}{\prod_{k=0}^{15} 2^k} 	imes \sum_{j=0}^{15} 2^{-j} = -2^{120} \sum_{j=0}^{15} 2^{-j} = -2^{120} (2 - 2^{-15}) = -2^{121} + 2^{105} = 2^{105} - 2^{121}$.

The coefficient of $x^{15}$ in the product $(1-x)(1-2x)(1-2^2x)(1-2^3x) \ldots (1-2^{15}x)$ is

Similar Questions

If the coefficient of $x$ in the expansion of $(ax^{2}+bx+c)(1-2x)^{26}$ is $-56$ and the coefficients of $x^{2}$ and $x^{3}$ are both zero, then $a+b+c$ is equal to:

Let $(1+x+x^2)^{2014} = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots + a_{4028} x^{4028}$. Let $A = a_0 - a_3 + a_6 - \ldots + a_{4026}$,$B = a_1 - a_4 + a_7 - \ldots - a_{4027}$,and $C = a_2 - a_5 + a_8 - \ldots + a_{4028}$. Then,

Let $(1+x+x^2)^{10}=a_0+a_1 x+a_2 x^2+\ldots+a_{20} x^{20}$. If $(a_1+a_3+a_5+\ldots+a_{19})-11 a_2=121 k$,then $k$ is equal to . . . . . . .

The sum of the series $2 \times 1 \times {}^{20}C_4 - 3 \times 2 \times {}^{20}C_5 + 4 \times 3 \times {}^{20}C_6 - 5 \times 4 \times {}^{20}C_7 + \dots + 18 \times 17 \times {}^{20}C_{20}$ is equal to:

Let $a > b > 0$ and $f(n) = a^{1/n} - b^{1/n}$, $J(n) = (a - b)^{1/n}$ for all $n \geq 2$. Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module