The sum of the coefficients of the terms of d | vedclass

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

Question

(D) The expansion is given by $(1+x)^n(1+y)^n(1+z)^n = \sum_{r=0}^n {}^nC_r x^r \sum_{s=0}^n {}^nC_s y^s \sum_{t=0}^n {}^nC_t z^t = \sum_{r,s,t} ({}^n

Vedclass · Accepted Answer

$({}^{3n}C_m)$

Vedclass · Answer

(D) The expansion is given by $(1+x)^n(1+y)^n(1+z)^n = \sum_{r=0}^n {}^nC_r x^r \sum_{s=0}^n {}^nC_s y^s \sum_{t=0}^n {}^nC_t z^t = \sum_{r,s,t} ({}^nC_r {}^nC_s {}^nC_t) x^r y^s z^t$.The terms of degree $m$ are those for which $r + s + t = m$,where $0 \le r, s, t \le n$.The sum of the coefficients of these terms is the sum of the products ${}^nC_r {}^nC_s {}^nC_t$ over all non-negative integers $r, s, t$ such that $r + s + t = m$.This is equivalent to the coefficient of $x^m$ in the expansion of $(1+x)^n(1+x)^n(1+x)^n = (1+x)^{3n}$.The coefficient of $x^m$ in $(1+x)^{3n}$ is given by ${}^{3n}C_m$.

The sum of the coefficients of the terms of degree $m$ in the expansion of $(1 + x)^n(1 + y)^n(1 + z)^n$ is:

Similar Questions

The coefficient of $x^6$ in the power series expansion of $\frac{x^4-12x^2+7}{(x^2+1)^3}$ is

Let $a$ be the largest real root and $b$ be the smallest real root of the polynomial equation $x^6-6x^5+15x^4-20x^3+15x^2-6x+1=0$. Then $\frac{a^2+b^2}{a+b+1}$ is

The sum of coefficients in the expansion of $(x + 2y + 3z)^8$ is

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:

If $x$ is so small that all terms containing $x^2$ and higher powers of $x$ can be neglected, then the approximate value of $\frac{(1+\frac{2x}{3})^{-4}(4+5x)^{1/2}}{(9+x)^{3/2}}$ when $x=\frac{6}{371}$, is

Vedclass Test Series

Exam Paper Generator

Online Exam Module