Respuesta :

Answer:

option-A

Step-by-step explanation:

We can use rth term binomial expansion formula

[tex](x+y)^n[/tex]

[tex]T_r=(n,r-1)x^{n-(r-1)}y^{r-1}[/tex]

we are given

[tex](a-\sqrt{2})^8[/tex]

we can compare

[tex]x=a,y=-\sqrt{2}[/tex]

n=8

r=4

now, we can find 4th term

[tex]T_r=(8,4-1)a^{8-(4-1)}(-\sqrt{2})^{4-1}[/tex]

now, we can simplify it

[tex]T_4=\frac{8!}{5!3!}\times -2\sqrt{2}a^5[/tex]

[tex]T_4=56\times -2\sqrt{2}a^5[/tex]

[tex]T_4=-112\sqrt{2}a^5[/tex]

Answer:

Option A. [tex]=-112a^{5}\sqrt{2}[/tex]

Step-by-step explanation:

from the given formula of binomial [tex](a+b)^{n}=a^{n}+na^{n-1}b+\frac{n(n-1)}{2!}a^{n-2}b^{2}+\frac{n(n-1)(n-2)}{3!}a^{n-3}b^{3}.....b^{n}[/tex] we can calculate any term of this expansion.

Now from the question we have to find out the 4th term of [tex](a-\sqrt{2} )^{8}[/tex]

From the expansion fourth term will be [tex]\frac{n(n-1)(n-2)}{3!}a^{n-3}b^{3}[/tex]

Now we replace a=a, b=(-√2) and n=8

[tex]\frac{8(8-1)(8-2)}{3!}a^{8-3}(-\sqrt{(2)})^{3}[/tex]

[tex]= -\frac{8\times 7\times 6}{3\times 2\times 1}a^{5}(2)\sqrt{2}[/tex][tex]=-8\times 7a^5(2)\sqrt{2}[/tex]

[tex]=-112a^{5}\sqrt{2}[/tex]


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