The number of terms in the expansion of (1 + | vedclass

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

Question

(C) The given expression is $(1 + x)^{101} (1 - x + x^2)^{100}$.We can rewrite this as $(1 + x) (1 + x)^{100} (1 - x + x^2)^{100}$.$= (1 + x) [(1 + x)

Vedclass · Accepted Answer

$202$

Vedclass · Answer

(C) The given expression is $(1 + x)^{101} (1 - x + x^2)^{100}$.We can rewrite this as $(1 + x) (1 + x)^{100} (1 - x + x^2)^{100}$.$= (1 + x) [(1 + x)(1 - x + x^2)]^{100}$.Using the identity $(a + b)(a^2 - ab + b^2) = a^3 + b^3$,we get $(1 + x)(1 - x + x^2) = 1 + x^3$.So,the expression becomes $(1 + x)(1 + x^3)^{100}$.The expansion of $(1 + x^3)^{100}$ contains $100 + 1 = 101$ terms.Multiplying by $(1 + x)$,we get $(1 + x)(1 + x^3)^{100} = (1 + x^3)^{100} + x(1 + x^3)^{100}$.Each part has $101$ terms,and since the powers of $x$ in $(1 + x^3)^{100}$ are multiples of $3$ $(0, 3, 6, \dots, 300)$ and the powers in $x(1 + x^3)^{100}$ are $1, 4, 7, \dots, 301$,there are no common terms.Therefore,the total number of terms is $101 + 101 = 202$.

The number of terms in the expansion of $(1 + x)^{101} (1 + x^2 - x)^{100}$ in powers of $x$ is

Similar Questions

Given $(1 - 2x + 5x^2 - 10x^3) (1 + x)^n = 1 + a_1x + a_2x^2 + \dots$ and that $a_1^2 = 2a_2$,then the value of $n$ is:

Let $R = (5\sqrt{5} + 11)^{2n + 1}$ and $f = R - [R]$,where $[.]$ denotes the greatest integer function. The value of $R \cdot f$ is

$\binom{n}{0} + 2\binom{n}{1} + 2^2\binom{n}{2} + \dots + 2^n\binom{n}{n}$ is equal to

The expression $\frac{1}{{\sqrt {4x + 1} }}\left[ {{{\left[ {\frac{{1 + \sqrt {4x + 1} }}{2}} \right]}^7} - {{\left[ {\frac{{1 - \sqrt {4x + 1} }}{2}} \right]}^7}} \right]$ is a polynomial in $x$ of degree

The value of $(\sqrt{5} + 1)^5 - (\sqrt{5} - 1)^5$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module