Question

Find the 4th term of the expansion A B C D

Answer 1

Answer:

Option A. $=-112a^{5}\sqrt{2}$

Step-by-step explanation:

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

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

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

Now we replace a=a, b=(-√2) and n=8

$\frac{8(8-1)(8-2)}{3!}a^{8-3}(-\sqrt{(2)})^{3}$

$= -\frac{8\times 7\times 6}{3\times 2\times 1}a^{5}(2)\sqrt{2}$$=-8\times 7a^5(2)\sqrt{2}$

$=-112a^{5}\sqrt{2}$

Answer 2

Answer:

option-A

Step-by-step explanation:

We can use rth term binomial expansion formula

$(x+y)^n$

$T_r=(n,r-1)x^{n-(r-1)}y^{r-1}$

we are given

$(a-\sqrt{2})^8$

we can compare

$x=a,y=-\sqrt{2}$

n=8

r=4

now, we can find 4th term

$T_r=(8,4-1)a^{8-(4-1)}(-\sqrt{2})^{4-1}$

now, we can simplify it

$T_4=\frac{8!}{5!3!}\times -2\sqrt{2}a^5$

$T_4=56\times -2\sqrt{2}a^5$

$T_4=-112\sqrt{2}a^5$