Question

Help please!

Answer 1

Let (2x^2+3x+4)^10=summation of arx^r (r goes from 0 to 20) then a7/a13=?

Answer 2

I know a way but it's quite a long procedure.We need to evaluate too much of factorial terms and then add them .Can anyone devise another way to get through this problem

Answer 3

@parthkohli

Answer 4

What is a7?

Answer 5

Does \(a_7=7ax^7\)?

Answer 6

I mean it is coefficient of x^7.

Answer 7

So what is the summation of arx^7 thing?

Answer 8

I mean summation of arx^r from 0 to 20.

Answer 9

It represents the expansion of (2x^2+3x+4)^10

Answer 10

arx^r? Do you mean ax^r? So you seek to expand the whole polynomial?

Answer 11

Yeah..and then equate the coefficients of that expansion with the summation part on RHS.
So you see we will have the values of a1,a2,a3,...so on.

Answer 12

Do you mean \(a_rx^r\) instead of \(a_rrx^r\)?

Answer 13

What methods do you have in mind?

Answer 14

I used computer to expand the polynomial. Although a7/a13 is a small integer there is no way you could calculate the expansion by hand. a7 is a 9 digit number...

Answer 15

I tried to solve it using this method.
{10!/(p!q!r!)}* (2x^2)^p(3x)^q(4)^r
Then p+q+r=10 and then randomly assuming the values which satisfy them such that p,q,r >=0

Answer 16

I am solving it this way.
\[
\left(2x^2+3x+4\right)^{10}=\sum_{k=0}^{10}\left(2x^2\right)^{k}\left(3x+10\right)^{10-k}
\]
For \(a_7\), k=1,2,3. For \(a_{13}\), k=3,4,5,6.

Answer 17

Correction: k=0,1,2,3 for \(a_7\)

Answer 18

Were about to prompt !

Answer 19

Isn't it too Long though?

Answer 20

Is that an exam question of some sort? I don't see how it can be done without computer assistance or a lot of patience.

Answer 21

Yes,it's a question of my exam that I didn't get .

Answer 22

It is actually doable by my method because of the massive cancellation of binomial coefficients.

Answer 23

Doable is okay with me.But is it too much of time taking solution

Answer 24

You are saying because you might be using a calC or something like that but i want a short way to have it solved.
Please can you help ?

Answer 25

How much time can be allocated to this question base on the marks given for this question?

Answer 26

The time allocated is not mentioned.

Answer 27

Well ,i randomly thought some way to do it though. You may want to have a glance it.

Answer 28

How about the marks allocated? Since your goal is to maximise the mark on your exam you can approximately deduce how much time you can allocate to one question before you must move on.

Answer 29

I think that i devote around 7-8 minutes for these type of questions .

Answer 30

Nvm..them infact they are literally great deal of time consuming questions for me atleast.
So will you help me in moving further with that method .

Answer 31

Yes?

Answer 32

It is like i replaced x by 1/x in the expansion.

Answer 33

Don't quite get it...

Answer 34

The answer is 8 as computed by computer. It took me 40 minutes to do so although I haven't eaten anything yet and I just woke up.

Answer 35

I can see if you know multinomial theorem you can do it much quicker using my idea.

Answer 36

Thanks, but i got it.