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

Let $(1 - 2x + 3x^2)^{10} = a_0 + a_1x + a_2x^2 + \dots + a_n x^n$,where $a_n \neq 0$. Then the arithmetic mean of $a_0, a_1, a_2, \dots, a_n$ is

The total number of terms in the expansion of $\left[ (1 + x)^{100} + (1 + x^2)^{100}(1 + x^3)^{100} \right]$ is

The value of $4 \{^nC_1 + 4 \cdot ^nC_2 + 4^2 \cdot ^nC_3 + \dots + 4^{n-1} \cdot ^nC_n\}$ is:

If the sum of the coefficients in the expansion of $(\alpha x^2 - 2x + 1)^{35}$ is equal to the sum of the coefficients in the expansion of $(x - \alpha y)^{35}$,then $\alpha = $

If $1 + x^4 + x^5 = \sum\limits_{i = 0}^5 a_i (1 + x)^i$ for all $x$ in $\mathbb{R},$ then $a_2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module