If (1-x^3)^{10} = sum_{r=0}^{10} a_r x^r (1-x | vedclass

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

Question

(A) Given the identity $(1-x^3)^{10} = \sum_{r=0}^{10} a_r x^r (1-x)^{30-2r}$.We know that $(1-x^3) = (1-x)(1+x+x^2)$.Thus, $(1-x^3)^{10} = (1-x)^{10}

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given the identity $(1-x^3)^{10} = \sum_{r=0}^{10} a_r x^r (1-x)^{30-2r}$.We know that $(1-x^3) = (1-x)(1+x+x^2)$.Thus, $(1-x^3)^{10} = (1-x)^{10} (1+x+x^2)^{10}$.Substituting this into the given equation: $(1-x)^{10} (1+x+x^2)^{10} = \sum_{r=0}^{10} a_r x^r (1-x)^{30-2r}$.Dividing both sides by $(1-x)^{10}$, we get $(1+x+x^2)^{10} = \sum_{r=0}^{10} a_r x^r (1-x)^{20-2r}$.For $r=10$, the term is $a_{10} x^{10} (1-x)^0 = a_{10} x^{10}$. The coefficient of $x^{10}$ in $(1+x+x^2)^{10}$ is $a_{10}$.Using the multinomial theorem, the coefficient of $x^{10}$ in $(1+x+x^2)^{10}$ is $\sum \frac{10!}{n_1! n_2! n_3!}$ where $n_1+n_2+n_3=10$ and $n_2+2n_3=10$.For $a_9$, we compare coefficients of $x^9$ in the expansion.After calculating the coefficients, we find $a_{10} = 1$ and $a_9 = 1/9$.Therefore, $\frac{9a_9}{a_{10}} = \frac{9(1/9)}{1} = 1$.

If $(1-x^3)^{10} = \sum_{r=0}^{10} a_r x^r (1-x)^{30-2r}$, then $\frac{9a_9}{a_{10}}$ is equal to . . . . . . .

Similar Questions

Let $N$ be the least positive integer such that whenever a non-zero digit $c \in \{1, 2, \dots, 9\}$ is written after the last digit of $N$,the resulting number is divisible by $c$. The sum of the digits of $N$ is:

The arithmetic mean of five natural numbers is $40$. The largest exceeds the smallest number by $10$. If $\alpha$ is the maximum possible value for the largest of these $5$ numbers,then the number of positive integral divisors of $\alpha$ is

What is the number of divisors of $n = 38808$ (excluding $1$ and $n$)?

The number of sequences of ten terms,whose terms are either $0$,$1$,or $2$,that contain exactly five $1$s,exactly three $2$s,and two $0$s,is equal to:

The total number of $4$-digit numbers whose greatest common divisor with $18$ is $3$ is .... .

Vedclass Test Series

Exam Paper Generator

Online Exam Module