For natural numbers m, n,if (1 - y)^m(1 + y)^ | vedclass

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

Question

(D) Given $(1 - y)^m(1 + y)^n = 1 + a_1y + a_2y^2 + \ldots$Expanding the binomials:$(1 - my + \frac{m(m-1)}{2}y^2 - \ldots)(1 + ny + \frac{n(n-1)}{2}y

Vedclass · Accepted Answer

$(35, 45)$

Vedclass · Answer

(D) Given $(1 - y)^m(1 + y)^n = 1 + a_1y + a_2y^2 + \ldots$Expanding the binomials:$(1 - my + \frac{m(m-1)}{2}y^2 - \ldots)(1 + ny + \frac{n(n-1)}{2}y^2 + \ldots) = 1 + a_1y + a_2y^2 + \ldots$Comparing the coefficient of $y$:$a_1 = n - m = 10 \implies n = m + 10$Comparing the coefficient of $y^2$:$a_2 = \frac{n(n-1)}{2} - nm + \frac{m(m-1)}{2} = 10$$n^2 - n - 2nm + m^2 - m = 20$$(n - m)^2 - (n + m) = 20$Since $n - m = 10$,we have $(10)^2 - (m + 10 + m) = 20$$100 - 2m - 10 = 20$$90 - 2m = 20$$2m = 70 \implies m = 35$$n = 35 + 10 = 45$Thus,$(m, n) = (35, 45)$.

For natural numbers $m, n$,if $(1 - y)^m(1 + y)^n = 1 + a_1y + a_2y^2 + \ldots$ and $a_1 = a_2 = 10$,then $(m, n) = \_\_\_\_\_\_$.

Similar Questions

Find the remainder when $3 \times 7^{22} + 2 \times 10^{22} - 44$ is divided by $18$.

Let $[x]$ denote the greatest integer less than or equal to $x$. If for $n \in N$,$(1-x+x^3)^n = \sum_{j=0}^{3n} a_j x^j$,then $\sum_{j=0}^{[\frac{3n}{2}]} a_{2j} + 4 \sum_{j=0}^{[\frac{3n-1}{2}]} a_{2j+1}$ is equal to

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

Expand the expression $(2x - 3)^6$.

If $n$ is a positive integer,then the coefficient of $x^6$ in the expansion of $(1-2x+3x^2-4x^3+\ldots)^{-n}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module