Respuesta :
Answer:
Part a: The coefficient of [tex]v^9w^2x^5y^7z^2[/tex] is [tex]1.766 \times 10^{13}[/tex]
Part b: The number of terms are 23751.
Step-by-step explanation:
part a
From the given equation the
[tex](x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx[/tex]
So the coefficient of any term [tex]x^ny^nz^n[/tex] is given as
[tex]\dfrac{2!}{n!n!n!}[/tex]
Similarly for the generic equation of coefficient of the term [tex]x_1^{r_1}x_2^{r_2}x_3^{r_3}......x_k^{r_k}[/tex] in the equation of form [tex](x_1+x_2+x_3+x_4......x_k)^n[/tex] is given as
[tex]\dfrac{n!}{r_1!r_2!r_3!....r_k!}[/tex]
So now the coefficient of [tex]v^9w^2x^5y^7z^2[/tex] is given as
[tex]=\dfrac{25!}{9!2!5!7!2!}\\=1.766 \times 10^{13}[/tex]
The coefficient of [tex]v^9w^2x^5y^7z^2[/tex] is [tex]1.766 \times 10^{13}[/tex]
Part b:
The number of terms is given as [tex]\left (\ {{m+n-1} \atop {n}} \right )[/tex]
where m is the number of variables which are 5 here
n is the power which is 25 so the number of variables is given as
[tex]\left (\ {{5+25-1} \atop {25}} \right )\\\left (\ {{29} \atop {25}} \right )\\\dfrac{29!}{25!4!}=23751[/tex]
So the number of terms are 23751.
Part(a): The coefficient of the term [tex]v^{9}w^{2}x^{5}y^{7}z^{2}[/tex] is [tex]\frac{25!}{9!\times 2!\times 5!\times 7!\times 2!}[/tex]
Part(b): The number terms are 23751
Part(a):
Given,
The given expression is, [tex](x^{2}+y^{2}+z^{2})^{2}[/tex]
The number of coeffecient is given by,
[tex]\frac{(9+2+5+7+2)!}{9!\times 2!\times 5!\times 7!\times 2!}} =\frac{25!}{9!\times 2!\times 5!\times 7!\times 2!}[/tex]
Part(b):
The given expression is, [tex](x^{2}+y^{2}+z^{2})^{2}[/tex]
Let, [tex]v^{9}w^{2}x^{5}y^{7}z^{2}[/tex] be the term of the expansion [tex](v+w+x+y+z)^{25}[/tex]
Therefore, [tex]a+b+c+d+e=25[/tex]
The total number of different terms is equal to the number of ways we can distribute 25 items among 5 people.
Then the number of different terms are,
[tex]^{25+5-1}C_{5-1}=^{29}C_{4}\\=23751[/tex]
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