Question

plz ans..fast



the coefficient of x^4 in the expansion of (1-2x)^5 is equal to.​

Answer 1

Answer:

80

Step-by-step explanation:

Given Expansion :

$\sf ( 1-2x)^5$

We know :

$\sf T_{r+1}= \ ^nC_r \ (a)^{n-r} \ (b)^r$

$\sf T_{r+1}= \ ^5C_r \ (1)^{5-r} \ (-2x)^r$

$\sf T_{r+1}= \ ^5C_r \ (-2x)^r \ x^r$

For coefficient of x⁴, power of x = 4

r = 4

Putting value of r = 4

$\sf \rightarrow {^5C_4 \ ( -2x)^4}$

$\sf \rightarrow {^5C_4 \ -16 x^4}$

Coefficient part :

= > ( 5! / 4! × 1! ) × 16

= > 5 × 16

= > 80 .

Therefore , we get required answer.