two problems in binomial theorem

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October 13th 2012, 01:10 AM
mido22

two problems in binomial theorem

1)In the expansion of $(2x-3)^1^5$ if $13T3 + 10T4 + T5 = 0$ where Tn is the term number n
find the values of X that satisfy the given
....My solution : x= 0.5 , 4.5 , 0
Guide answer : 0.5 , 4.5 only ???

2)If the ratio between the two middle terms in the expansion of $(1+x)^2^7$ is 1:4 find the value of X
My solution : x=4 or 0.25
Guide answer : 4 only ???
October 13th 2012, 02:43 AM
chiro

Re: two problems in binomial theorem

Hey mido22.

Can you show your working for the problems?
October 13th 2012, 03:20 AM
mido22

Re: two problems in binomial theorem

1) $t3=945 (2x)^1^3 , t4=-12285(2x)^1^2 , t5=110565(2x)^1^1$ --> from binomial theorem

since $13 T3 +10T4 + T5 =0$

so $12285 (2x)^1^1 (4x^2 -20x +9) =0$

so x= zero or 0.5 or 4.5 since $2x^1^1 =0$ or $4x^2-20x+9=0$
October 13th 2012, 03:24 AM
mido22

Re: two problems in binomial theorem

2) $t14 : t15 = 1:4 or 4:1$ because he didn't say in the problem the between t14 to t15 or vice versa

$27C13 . x^1^4 : 27C14 . x^1^3 =1:4 or 4:1$

$so 27C13 . x^1^4 : 27C13 . x^1^3 =1:4 or 4:1$

$so x:1 = 1:4 or 4:1$

$so x=4 or 0.25$
October 13th 2012, 04:47 PM
chiro

Re: two problems in binomial theorem

For 1) I don't know why they didn't have 0 as a solution: you're answer looks spot on.

For 2) I think they mean successive terms so by this, they mean that the next term is 4 times greater than the previous one.

So basically divide T14/T13 = 4/1 = 4 which gives x = 4.

This is just my interpretation though.

You should ask your teacher for clarification, especially for number 1 in which you gave a good answer. Even your answer for question 2 is a solid one as well.
October 13th 2012, 05:43 PM
Soroban

Re: two problems in binomial theorem

Hello, mido22!

Quote:

2) If the ratio between the two middle terms in the expansion of $(1+x)^{27}$ is $1:4$
find the value of $x.$

The two middle terms are the 14th and 15th: . ${27\choose13}x^{13}\text{ and }{27\choose14}x^{14}$

The ratio is: . $\frac{{27\choose13}x^{13}}{{27\choose14}x^{14}} \:=\:\frac{1}{4} \quad\Rightarrow\quad \frac{1}{x} \:=\:\frac{1}{4} \quad\Rightarrow\quad x \:=\:4$

I assumed that the ratio was in that order: 14th to the 15th.
But I agree that the wording could have been more specific.

I would have allowed full credit (maybe more) for your answer.
It certainly demonstrates innovative thinking.
October 14th 2012, 04:04 AM
mido22

Re: two problems in binomial theorem

Thanks (Soroban and chiro) very much