Question

what is the 4th term of (a- √2)^8?

choice are a.112 √2a^4 b.-112 √2a^4
c.112 √2a^5 d.-112 √2a^5

please i need help

Answer 1

Do you know the nth term of a binomial series?

Answer 2

nope

Answer 3

i never learned this im a senior but i took this class last year and dont remember much.

Answer 4

hint :
the 4th term of (p - q)^n is nC3 p^(n-3) * (-q)^3

Answer 5

i dont understand can you further explain. sorry

Answer 6

The nth term of a binomial series (p - q)^n is
nC0 p^(n-0) * (-q)^0 + nC1 p^(n-1) * (-q)^1 + nC2 p^(n-2) * (-q)^2 + ... + nC3 p^(n-n) * (-q)^n

look at the binomial series above :
(a- √2)^8. it means p =a, q = -√2, n = 8. therefore the 4th term of (a- √2)^8 is 8C3 a^(8-3) * (-√2)^3 then evaluate all values here.

use the combination formula to get 8C3.
nCr = n!/(n-r)!r!

Answer 7

wlcm.
let's see what you get then ...

Answer 8

@11calcBC

Answer 9

oh wtf it a-root2...

Answer 10

it's*

Answer 11

u could just solve it algebraically by solving for a numerical value, then check it

Answer 12

oh okay

Answer 13

lol sry i would use a better method but i gtg get dinner so..sry mate :D

Answer 14

RadEn gave you the formula above. The fourth term is: 8C3 a^(8-3) * (-√2)^3

Let us go one term at a time. What do you get for 8C3?