If in the expansion of (1 + x)^m(1 - x)^n,the | vedclass

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

Question

(C) The expansion is given by $(1 + x)^m(1 - x)^n = (1 + mx + \frac{m(m - 1)}{2}x^2 + \dots)(1 - nx + \frac{n(n - 1)}{2}x^2 - \dots)$.Multiplying the

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(C) The expansion is given by $(1 + x)^m(1 - x)^n = (1 + mx + \frac{m(m - 1)}{2}x^2 + \dots)(1 - nx + \frac{n(n - 1)}{2}x^2 - \dots)$.Multiplying the terms,we get $1 + (m - n)x + [\frac{n(n - 1)}{2} - mn + \frac{m(m - 1)}{2}]x^2 + \dots$.Given the coefficient of $x$ is $3$,we have $m - n = 3$,so $n = m - 3$.Given the coefficient of $x^2$ is $-6$,we have $\frac{n^2 - n}{2} - mn + \frac{m^2 - m}{2} = -6$.Substituting $n = m - 3$ into the equation:$\frac{(m - 3)(m - 4)}{2} - m(m - 3) + \frac{m^2 - m}{2} = -6$.Multiplying by $2$: $(m^2 - 7m + 12) - 2(m^2 - 3m) + (m^2 - m) = -12$.$m^2 - 7m + 12 - 2m^2 + 6m + m^2 - m = -12$.$-2m + 12 = -12$.$-2m = -24$,which gives $m = 12$.

If in the expansion of $(1 + x)^m(1 - x)^n$,the coefficients of $x$ and $x^2$ are $3$ and $-6$ respectively,then $m$ is:

Similar Questions

Using the Binomial Theorem,evaluate $(99)^{5}$.

Let $n > 2$ be an integer and define a polynomial $p(x) = x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0$,where $a_0, a_1, \ldots, a_{n-1}$ are integers. Suppose we know that $n p(x) = (1 + x) p'(x)$. If $b = p(1)$,then:

Find the value of $\frac{18^3 + 7^3 + 3 \times 18 \times 7 \times 25}{3^6 + 6 \times 243 \times 2 + 15 \times 81 \times 4 + 20 \times 27 \times 8 + 15 \times 9 \times 16 + 6 \times 3 \times 32 + 64}$

Expand the expression $\left(\frac{2}{x}-\frac{x}{2}\right)^{5}$

The sum of the coefficients in the expansion of $(1+x+x^2)^n$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module