Question

Given that:

(1+x)^5 (1+ax)^6 ≡ 1 + bx + 10x^2 + ..... + a^6x^11,

find the values of a,b∈Z

I'm not entirely sure how to go about doing this. Please help? :)

Answer 1

1. use the binomial expansion to expand each bracket up to the the terms involving \(x^2\)
2. then multiply these two keeping only the terms up to \(x^2\) (i.e. ignore any terms of higher powers of x)
3. then compare the result with \(1 + bx + 10x^2\) and you should get some equations involving a and b that can be solved.

Answer 2

Thanks I'll do the expansion and see if I'm on the right track.

Answer 3

great! let me know if you get stuck.

Answer 4

when I do the expansion, I get = 1+(30ax^2)+(150*a^2*x^4)

Am I on the right track? thanks

Answer 5

what is this the expansion for?

Answer 6

for (1+x)^5 (1+ax)^6

Answer 7

That does not look correct.
I suggest you do this step-by-step.
first show what you believe will be the expansion for (1+x)^5
then the expansion for (1+ax)^6
then we can work on the product

Answer 8

Yeah, it seems I've done too many steps at once and got confused. I'll work on them separately. :)

Answer 9

I get:

(1+x)^5 = 1 + 5x + 10x^2

(1+ax)^6 = 1 + 6ax + 15ax^2

Answer 10

I /think/ the 3rd term in the 2nd expansion should have \(a^2\) in it

Answer 11

i.e.:\[(1+ax)^6 = 1 + 6ax + 15a^2x^2\]

Answer 12

Ah yse, forgot the brackets.

Answer 13

think of "ax" as one quantity

Answer 14

ok, good, now you need to calculate this beast:\[(1 + 5x + 10x^2)(1 + 6ax + 15a^2x^2)\]

Answer 15

Should I do it term by term?

Answer 16

first multiply the 1 by each of the three terms, then the 5x ... etc.?

Answer 17

yes - I would break it up into three parts as follows:\[1(1 + 6ax + 15a^2x^2)+5x(1 + 6ax + 15a^2x^2)+10x^2(1 + 6ax + 15a^2x^2)\]

Answer 18

and remember to throw away all powers of x greater than 2

Answer 19

ah so I'm allowed to just disregard them? That's pretty handy. :D

Answer 20

yes :)

Answer 21

great thanks :)

Answer 22

yw :)

Answer 23

throwing away the powers above x^2, I get the product to be:

\[1 + 6ax + 15a ^{2}x ^{2}+5x+30ax ^{2}+10x ^{2}\]

Answer 24

good - now gather like terms and you almost there.

Answer 25

\[1 + (6a+5)x + (15a ^{2}+30a+10)x ^{2}\]

so b = 6a+5 and the other bracket equals to 10?

Answer 26

correct

Answer 27

you should get two solutions for 'a'

Answer 28

I get a = (b-5)/6 and a = -2

Answer 29

oh, I plug -2 into the first a equation to get b?

Answer 30

a=-2 is correct but you missed another value for a.
once you know the values for a, plug them into b=6a+5 to get the values for b.

Answer 31

I got that a(a+2) = 0 so a = 0 or -2, but since a belongs to Z then it's just -2? Or is 0 a valid value too then?

Answer 32

I thought Z included all negative, zero, and positive numbers?

Answer 33

so a=0 is a solution as well

Answer 34

Ah yes, probably. Thanks very much for the help! To think this isn't even the toughest question on my assigment. ... -.- :)

Answer 35

yw :)
and good luck with the rest!

Answer 36

Cheers!

Answer 37

np my friend! :)