How to find greatest coefficient in an expansion? (1 Viewer)

[nabzilla]

This is a binomial question, and I'm having lots of trouble. Say if the expansion was (3+4x)^16 how would I find the greatest coefficient? Do I just expand it all? Or is there an actual formula?

 

[Bored Of Fail 2]

This is a binomial question, and I'm having lots of trouble. Say if the expansion was (3+4x)^16 how would I find the greatest coefficient? Do I just expand it all? Or is there an actual formula?

theres a formula

T (k+1) / T (k) < 1 im pretty sure if memory serves me right

it basically means that when the coefficents of the terms FIRST start to decrease, then you have reached the greatest term

 

[nabzilla]

theres a formula

T (k+1) / T (k) < 1 im pretty sure if memory serves me right

it basically means that when the coefficents of the terms FIRST start to decrease, then you have reached the greatest term

And then do I solve for k and sub back into equation or something? Or do I just leave it as k<whatever comes up?

 

[Bored Of Fail 2]

http://img121.imageshack.us/i/binomialh.jpg/

see this link for working out

 

[Bored Of Fail 2]

so we have to solve 3 (17-k) / 4k < 1

now k>0 so multiply both sides by 4k


3 ( 17-k) < 4k
51 -3k < 4k
7k > 51
k> 7.28 .....

but k is an integer, so take the first integer value of k that satisifes this

so k= 8

ie the greatest coefficicent is the 9th term ( k starts at 0 )

now sub this into the T (k+1) formula

so answer = ( 16 C 8 ) ( 3^8)(4^8)

 

[nabzilla]

http://img121.imageshack.us/i/binomialh.jpg/

see this link for working out

Omg wow. Thank you so much for your help!!

 

[nabzilla]

so we have to solve 3 (17-k) / 4k < 1

now k>0 so multiply both sides by 4k


3 ( 17-k) < 4k
51 -3k < 4k
7k > 51
k> 7.28 .....

but k is an integer, so take the first integer value of k that satisifes this

so k= 8

ie the greatest coefficicent is the 9th term ( k starts at 0 )

now sub this into the T (k+1) formula

so answer = ( 16 C 8 ) ( 3^8)(4^8)

Oh that makes so much more sense! Thank you so much!!

 

[Bored Of Fail 2]

have you got the actual answer to it, im not 100% if what I came up with is correct, I subbed in other values of k and it didnt work out to be the largest coefficent.

e.g. ( 16C6 ) ( 3^6) ( 4^10 ) is bigger than (16C8) ( 3^8) ( 4^8)

im not quite sure where I when wrong.

do you have the actual answer??

 

[nabzilla]

Nope, I don't have the actual answer.

Wait hang on, wouldn't it be 16ck x 3^16-k x 4^k / 16c(k-1) x 3^17-k x 4^k-1 ?

 

[Bored Of Fail 2]

AHHHH yes, thats where the mistake is!!

yeh, but everything elsen is the same kind of method

you just have to get used to simplify the divisions of nCk and nC ( k-1)

and knowing how to express k! as (k-1)! x k ]

yeh, just fix up the working out and the answer should come out lol

 

[nabzilla]

haha alright. Thank you for your help!

 

[Bored Of Fail 2]

nah, the inequality was wrong, it should be 3 (17-k) / 4k > 1



so 51 -3k > 4k
7k < 51
k< 7.28...

therefore k = 7

so sub into the T (k+1) formula , answer = (16C7) ( 3^7) ( 4^9) , and checking the coefficents either end of it confirms that this is indeed the maxium

I havent really done alot of this in the last yr, they never touch this stuff in uni math

 

[Bored Of Fail 2]

(16C7) ( 3^7) ( 4^9)

^I promise that is the answer , always a good idea to check the coefficents of the terms to the left and the right of the answer you come up with, can help weed out noob mistakes like I did.

 

[Cardscook77]

see this link for working out

do you have an updated link for the working? I am also quite confused.

 

[yanujw]

do you have an updated link for the working? I am also quite confused.

The process + an example is explained in depth here