The coefficient of x^{1012} in the expansion | vedclass

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

Question

(B) The given expression is $(1 + x^n + x^{253})^{10}$.Using the multinomial theorem,the general term is given by $\frac{10!}{a!b!c!} (1)^a (x^n)^b (x

Vedclass · Accepted Answer

$^{10}C_4$

Vedclass · Answer

(B) The given expression is $(1 + x^n + x^{253})^{10}$.Using the multinomial theorem,the general term is given by $\frac{10!}{a!b!c!} (1)^a (x^n)^b (x^{253})^c$,where $a + b + c = 10$.We need the coefficient of $x^{1012}$,so we set the exponent of $x$ to $1012$:$nb + 253c = 1012$.Since $c \leq 10$ and $253 	imes 4 = 1012$,we test values for $c$:If $c = 4$,then $nb = 1012 - 253(4) = 0$. Since $n$ is a positive integer,$b$ must be $0$.Then $a = 10 - b - c = 10 - 0 - 4 = 6$.The coefficient is $\frac{10!}{6!0!4!} = ^{10}C_4$.If $c Thus,the only solution is $a=6, b=0, c=4$.The coefficient is $^{10}C_4$.

The coefficient of $x^{1012}$ in the expansion of $(1 + x^n + x^{253})^{10}$,where $n \leq 22$ is any positive integer,is

Similar Questions

The coefficient of $a^{3} b^{4} c^{5}$ in the expansion of $(bc + ca + ab)^{6}$ is

The number of odd positive divisors of $67500$ is

The number of positive even divisors of $6300$ is

The number of positive divisors of $1080$ is

The number of positive integers $n$ in the set $\{2, 3, \ldots, 200\}$ such that $\frac{1}{n}$ has a terminating decimal expansion is

Vedclass Test Series

Exam Paper Generator

Online Exam Module