Question

What is the fourth term in the expansion of (x + 2)6?

Answer 1

U have to apply binomial theorem to expand this.....
do u know binomial theorem ?

Answer 2

no :/ but my options are:
480x

168x^4

160x^3

120x^3

Answer 3

options are pointless without knowing how to pick one

Answer 4

the powers descend from: 6 5 4 3, so 4th term needs to be a power of 3

Answer 5

oh, so it has to be either C or D?

Answer 6

that does narrow the options yes

Answer 7

Ya. right @amistre64 if u don't know how to expand , then options are useless....But in my view.. the fourth term will be 40x^3..... can anybody tel me where I made the mistake.... :(

Answer 8

I'm thinking it's D....?

Answer 9

i like to set up the general stuff like this; using the power, lets layout the first and last terms

1 2 3 4 5 6 7 term
x^6 x^5 x^4 x^3 x^2 x^1 x^0 firsts
2^0 2^1 2^2 2^3 2^4 2^5 2^6 lasts

so this thing is at least 2^4 x^3 but there is a leading coeff that would have to be worked out

Answer 10

6C0
6C1
6C2
6C3 <--- thats the one

Answer 11

6.5.4
---- = 20
3.2

20*2^4*x^3

Answer 12

so it is D? :)

Answer 13

2^3 .... 8*20 = 160

Answer 14

waleed; 20*2 was what you did, instead of 20*2^3

Answer 15

1 6 15 20 15 6 1 coeffs
x^6 x^5 x^4 x^3 x^2 x^1 x^0 firsts
2^0 2^1 2^2 2^3 2^4 2^5 2^6 lasts

Answer 16

ohh, hrm, so it's 160x^3, right

Answer 17

yes :)

Answer 18

I remember the binomial theorem as this... and it is the transformed form of the theorem without combinations.....
it is given below......
(a+b)^n =a^n +a^n-1(b)+n(n-1)/2{a^n-2(b)}^2+n(n-1)(n-2)/6{a^n-3(b)^3}...............+b^n

so this is the fourth term..n(n-1)(n-2)/6{a^n-3(b)^3} right
put a=x and b=2 and n=6.....

Answer 19

yes

Answer 20

ya. i got it...... 160x^3 is the answer ..I did wrong calculation last time but the formula was the sam.e

Answer 21

thank you!

Answer 22

My pleasure......

Answer 23

http://www.wolframalpha.com/input/?i=expand+%282%2Bx%29^6