The coefficient of x^{50} in the expansion of | vedclass

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

Question

(A) Given expression: $(1+x)^{101}(1-x+x^2)^{100}$ $= (1+x) \cdot (1+x)^{100} \cdot (1-x+x^2)^{100}$ $= (1+x) \cdot [(1+x)(1-x+x^2)]^{100}$ $= (1+x)(1

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Given expression: $(1+x)^{101}(1-x+x^2)^{100}$ $= (1+x) \cdot (1+x)^{100} \cdot (1-x+x^2)^{100}$ $= (1+x) \cdot [(1+x)(1-x+x^2)]^{100}$ $= (1+x)(1+x^3)^{100}$ $= (1+x^3)^{100} + x(1+x^3)^{100}$ We need the coefficient of $x^{50}$. In the expansion of $(1+x^3)^{100}$,the powers of $x$ are of the form $3k$,where $k$ is an integer. Since $50$ is not a multiple of $3$,the coefficient of $x^{50}$ in $(1+x^3)^{100}$ is $0$. Similarly,for $x(1+x^3)^{100}$,we need the coefficient of $x^{49}$ in $(1+x^3)^{100}$. Since $49$ is not a multiple of $3$,the coefficient of $x^{49}$ in $(1+x^3)^{100}$ is $0$. Therefore,the coefficient of $x^{50}$ is $0 + 0 = 0$.

The coefficient of $x^{50}$ in the expansion of $(1+x)^{101}(1-x+x^2)^{100}$ is

Similar Questions

In the expansion of ${\left( {3x - \frac{1}{{{x^2}}}} \right)^{10}}$,the $5^{th}$ term from the end is:

If $a_r$ is the coefficient of $x^{10-r}$ in the Binomial expansion of $(1+x)^{10}$,then $\sum \limits_{r=1}^{10} r^3\left(\frac{a_r}{a_{r-1}}\right)^2$ is equal to

The coefficient of $x^5$ in the expansion of $(1+x^2)^5(1+x)^4$ is:

The coefficient of $x^9$ in the polynomial given by $\sum_{r=1}^{11} {(x+r)(x+r+1)(x+r+2)...(x+r+9)}$ is

If $a^3 + b^6 = 2$,then the maximum value of the term independent of $x$ in the expansion of $(ax^{1/3} + bx^{-1/6})^9$ is,where $(a > 0, b > 0)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module