+0  
 
0
524
2
avatar

Find the coefficient of x^2 y^2 z^3 in the expansion of (4x + 5y - 3z)^7.

 May 18, 2020
 #1
avatar
0

16384 x^7 + 143360 x^6 y - 86016 x^6 z + 537600 x^5 y^2 - 645120 x^5 y z + 193536 x^5 z^2 + 1120000 x^4 y^3 - 2016000 x^4 y^2 z + 1209600 x^4 y z^2 - 241920 x^4 z^3 + 1400000 x^3 y^4 - 3360000 x^3 y^3 z + 3024000 x^3 y^2 z^2 - 1209600 x^3 y z^3 + 181440 x^3 z^4 + 1050000 x^2 y^5 - 3150000 x^2 y^4 z + 3780000 x^2 y^3 z^2 - 2268000 x^2 y^2 z^3 + 680400 x^2 y z^4 - 81648 x^2 z^5 + 437500 x y^6 - 1575000 x y^5 z + 2362500 x y^4 z^2 - 1890000 x y^3 z^3 + 850500 x y^2 z^4 - 204120 x y z^5 + 20412 x z^6 + 78125 y^7 - 328125 y^6 z + 590625 y^5 z^2 - 590625 y^4 z^3 + 354375 y^3 z^4 - 127575 y^2 z^5 + 25515 y z^6 - 2187 z^7
(36 terms)

 May 18, 2020
 #2
avatar+26367 
+1

Find the coefficient of \(x^2 y^2 z^3\) in the expansion of \((4x + 5y - 3z)^7\).

 

Multinomial theorem:

For any positive integer m and any nonnegative integer n,

the multinomial formula tells us how a sum with m terms expands when raised to an arbitrary power n:

\(\left( x_1+x_2+\cdots +x_m \right)^n = \sum \limits_{k_1+k_2+\dots+k_m=n} \dbinom{n}{k_1,k_2,\dots,k_m}\cdot x_1^{k_1}\cdot x_2^{k_2}\dots x_m^{k_m}\)

where

 

\(\dbinom{n}{k_1,k_2,\dots,k_m} = \dfrac{n!}{k_1!k_2!\dots k_m!} \)

 

\(\begin{array}{|lcll|} \hline x^2 y^2 z^3 \\ n=7,\ m=3 \\ \hline k_1 = 2,\ k_2 = 2,\ k_3 = 3 \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline (4x + 5y - 3z)^7 &=& \dots + \dfrac{7!}{2!2!3!} (4x)^2(5y)^2(-3z)^3 + \dots \\ \hline && \dfrac{7!}{2!2!3!} (4x)^2(5y)^2(-3z)^3 \\\\ &=& \dfrac{7!}{2!2!3!}* 16x^225y^2(-27)z^3 \\\\ &=& \dfrac{4*5*6*7}{4}* 16*25*(-27)x^2y^2z^3 \\\\ &=& -210* 10800*x^2y^2z^3 \\\\ &=& \mathbf{-2268000}x^2y^2z^3 \\\\ \hline \end{array}\)

 

Source: https://en.wikipedia.org/wiki/Multinomial_theorem

 

laugh

 May 19, 2020

4 Online Users

avatar
avatar
avatar