Finding the greatest term in binomial expansions HELP NEEDED! (2 Viewers)

blackops23 · Oct 11, 2011

Hi guys, new to binomials here, just wondering what you would do in these types of questions:

1. Find the greatest term in (1+3x)^7 if x = 2/5

2. Find the greatest coefficient in the expansion of (1+ (2/3)x)^10

So what do you do with these? Can someone please provide a solution of the quickest solution?

Thank you, appreciate the help!

Drongoski · Oct 11, 2011

Greatest term occurs when: $\frac {T_{i+1}}{T_i} \geq 1$

i.e. when: $\frac {\binom 7 {i+1} (3x)^{i+1}} {\binom 7 i(3x)^i} \geq 1$

b3kh1t · Oct 11, 2011

Drongoski said:
Greatest term occurs when: $\frac {T_{i+1}}{T_i} \geq 1$

i.e. when: $\frac {\binom 7 {i+1} (3x)^{i+1}} {\binom 7 i(3x)^i} \geq 1$

The method I use would be the same for either of the questions. However the only difference is for the greatest term you are given the x value so you substitue it into the equation and get $(1+3(\frac{2}{5}))^{7}$

$=(1+\frac{6}{5})^{7}$

You do not solve though, you then write it in the form Drongoski said, so $\frac{T_{k+1}}{T_{k}}\geq 1$ and you solve from there.

Attempt the question now and let me know if you have further troubles.

someth1ng · Oct 11, 2011

b3kh1t said:
$\frac{T_{k+1}}{T_{k}}\geq 1$

I never understood this. If Tk+1/Tk is >= 1, couldn't it be alot of things?

Drongoski · Oct 11, 2011

Some books derive a standard formula for the ratio:

$\frac {T_{k+1}}{T_k} = \frac {n-k+1}{k} \times \frac {b}{a}$

which, if allowed, simplifies your work somewhat.

hup · Oct 11, 2011

someth1ng said:
I never understood this. If Tk+1/Tk is >= 1, couldn't it be alot of things?

no because after the maximum term the terms will fall in value

Drongoski · Oct 11, 2011

someth1ng said:
I never understood this. If Tk+1/Tk is >= 1, couldn't it be alot of things?

You find the largest k satisfying the inequality.

someth1ng · Oct 11, 2011

Drongoski said:
You find the largest k satisfying the inequality.

Right, so you just go up until you get to the highest then it will stop satisfying the inequation?

b3kh1t · Oct 11, 2011

someth1ng said:
I never understood this. If Tk+1/Tk is >= 1, couldn't it be alot of things?

$\frac{T_{k+1}}{T_{k}}> 1$ can then be changed into

$T_{k+1}> T_{k}$

all that means is the next term is greater then the term before it.

math man · Oct 11, 2011

someth1ng said:
I never understood this. If Tk+1/Tk is >= 1, couldn't it be alot of things?

What this is actually trying to work out is which term (k+1) is greater than every other term k, the reason for the $\geq$ is cause there can sometimes be two greatest terms such in the expansion of

$(1+x)^{3}= 1 + 3x + 3x^{2} + x^{3}$

we see here that there are two greatest terms, $T_{2}, T_{3}$ and this actually will be the case for all odd powered expansions of just (1 +x) for n>0.

So in this case how would we determine if there are two greatest terms?

Well what would usually happen is that for:

$\frac{T_{k+1}}{T_{k}}\geq 1$

you would get:

$k\leq M\: \: or\: \: k\geq M$

where M is a positive integer. When you get an integer in the equality you must test the number M, then either the number one greater than for or the number one less than M depending on the inequality.

I hope this helps you understand the greatest coef

math man · Oct 11, 2011

And to actually answer this question:

We want to find the greatest coef of the expansion, and the greatest coef is given by:

$\frac{T_{k+1}}{T_{k}}\geq 1$

Now some tests actually make you derive the formula for the greatest coef, so ill do that here:

First remember the general term $T_{k+1}$ is given by:

$T_{k+1}= _{k}^{n}\textrm{C}(a)^{n-k}(b)^{k}\: \: for\: \: (a+b)^{n}$

Therefore, $T_{k}= _{k-1}^{n}\textrm{C}(a)^{n-k+1}(b)^{k-1}\: \: for\: \: (a+b)^{n}$

Equating the $\frac{T_{k+1}}{T_{k}}$ part becomes:

$\frac{T_{k+1}}{T_{k}}= \frac{_{k}^{n}\textrm{C}(a)^{n-k}(b)^{k}}{_{k-1}^{n}\textrm{C}(a)^{n-k+1}(b)^{k-1}}$

Using the definition for combinations and indice laws this simplifies to:

$\frac{T_{k+1}}{T_{k}}= \frac{\frac{n!}{k!(n-k)!}b}{\frac{n!}{(k-1)!(n-k+1)!}a}$

Then we simplify this further to:

$\frac{T_{k+1}}{T_{k}}= \frac{(k-1)!(n-k+1)!b}{k!(n-k)!a}$

We further simply this again to:

$\frac{T_{k+1}}{T_{k}}= \frac{(k-1)!(n-k+1)(n-k)!b}{k(k-1)!(n-k)!a}$

And finally we arrive at:

$\frac{T_{k+1}}{T_{k}}= \frac{(n-k+1)}{k}\frac{b}{a}$

Now we can use this formula to find the greatest term in your question:

$(1+3x)^{7}, x=\frac{2}{5}$

Now we have:

n=7

a=1

$b=\frac{6}{5}$

Our aim is to find k, so we can work out the term, $T_{k+1}$ , which has the greatest coef.

So first we sub these into the formula i derived above:

$\frac{T_{k+1}}{T_{k}}= \frac{(n-k+1)}{k}\frac{b}{a}$

and solve:

$\frac{(n-k+1)}{k}\frac{b}{a}\geq 1$

Subbing are above values this becomes:

$\frac{(7-k+1)}{k}\frac{\frac{6}{5}}{1}\geq 1$

and now we solve this for k:

$6(7-k+1)\geq 5k$

$48-6k\geq 5k$

$48\geq 11k$

and we get:

$k\leq \frac{48}{11}\approx 4.36$

Therefore this means k=4 will give us our greatest coef, and that means $T_{5}$ is the greatest coef, which im sure you can evaluate now

blackops23 · Oct 11, 2011

math man said:
So in this case how would we determine if there are two greatest terms?

Well what would usually happen is that for:

$\frac{T_{k+1}}{T_{k}}\geq 1$

you would get:

$k\leq M\: \: or\: \: k\geq M$

where M is a positive integer. When you get an integer in the equality you must test the number M, then either the number one greater than for or the number one less than M depending on the inequality.

I hope this helps you understand the greatest coef

Thankyou for the help math man, but I just have a few questions: On mathsonline.com.au, the guy in the video said you do T(k+1)/T(k) > 1 --> that is there is no inequality... I did a few exercises from the website and there were times where i got k < 6 --> i.e. a whole integer - and so I made k = 5, so T(k+1) = T(6) --> which turned out to be the CORRECT ANSWER --> basically I did not include the equality sign and use k=6, and yet the answer was still right

So do I need to include the inequality? The example you used to illustrate the occurrence of two greatest terms was (1+x)^3 --> but correct me if I am wrong, but would the occurrence of two greatest terms only occur if it was (1+x)^n, where n is odd?

Thank you, please explain to your best ability, thank you very much, appreciate the help!

someth1ng · Oct 11, 2011

math man said:
and we get:

$k\leq \frac{48}{11}\approx 4.36$

Therefore this means k=4 will give us our greatest coef, and that means $T_{5}$ is the greatest coef, which im sure you can evaluate now

How do you know it is four? Is it just rounding it off?

nightweaver066 · Oct 11, 2011

someth1ng said:
How do you know it is four? Is it just rounding it off?

k = 4 as it is less than 4.36 and k must be an integer.

someth1ng · Oct 11, 2011

nightweaver066 said:
k = 4 as it is less than 4.36 and k must be an integer.

Oh, didn't see that cheers.

blackops23 · Oct 11, 2011

blackops23 said:
Thankyou for the help math man, but I just have a few questions: On mathsonline.com.au, the guy in the video said you do T(k+1)/T(k) > 1 --> that is there is no inequality... I did a few exercises from the website and there were times where i got k < 6 --> i.e. a whole integer - and so I made k = 5, so T(k+1) = T(6) --> which turned out to be the CORRECT ANSWER --> basically I did not include the equality sign and use k=6, and yet the answer was still right

So do I need to include the inequality? The example you used to illustrate the occurrence of two greatest terms was (1+x)^3 --> but correct me if I am wrong, but would the occurrence of two greatest terms only occur if it was (1+x)^n, where n is odd?

Thank you, please explain to your best ability, thank you very much, appreciate the help!

someone please answer my question LOL thanks

nightweaver066 · Oct 11, 2011

blackops23 said:
someone please answer my question LOL thanks

You must use $\geq$ as you end up with exactly the same two terms but them being the greatest.

For your question, you ended up with $k \leq 6$ so k = 6. You end up with it being 2 terms that are equal.

If it were $k \leq 3.24$ , you must take $k = 3$ and end up with only 1 term.

Remember the condition you used,
$T_{k+1}\geq T_k$ if the ratio between them two is $\geq 1$
As you take a whole integer that is below 3.24, it becomes $T_{k+1} > T_k$ . If you could somehow take 3.24 (but you cannot in a question), it would be $T_{k+1} = T_k$

So back to what you ended up with, $k \leq 6$

So you take the integer k = 6, and since T(k+1) = T(k), T(7) = T(6) and so you can evaluate either of those and end up with the greatest term.

Also, it does not matter if n is odd or not. Not all binomials given to you are in the form of (1 + ax)^n. They can be $(\frac{15}{x^2} - x^3)^{12}$ . If a binomial was in the form of (1 + x)^n where n is odd, you would end up with 2 term being the same and greater than all other terms.

math man · Oct 11, 2011

blackops23 said:
Thankyou for the help math man, but I just have a few questions: On mathsonline.com.au, the guy in the video said you do T(k+1)/T(k) > 1 --> that is there is no inequality... I did a few exercises from the website and there were times where i got k < 6 --> i.e. a whole integer - and so I made k = 5, so T(k+1) = T(6) --> which turned out to be the CORRECT ANSWER --> basically I did not include the equality sign and use k=6, and yet the answer was still right

So do I need to include the inequality? The example you used to illustrate the occurrence of two greatest terms was (1+x)^3 --> but correct me if I am wrong, but would the occurrence of two greatest terms only occur if it was (1+x)^n, where n is odd?

Thank you, please explain to your best ability, thank you very much, appreciate the help!

Yeh the expansion of (1+x) to the n, if n is odd will always have two greatest coef's as i did say before. And for your above example, sub k=6 in and see if it is the same answer, then get back to me
The reason for the equality part is there can be two greatest terms as i said before, so subbing k=5 is right, but sub k=6 and tell me what you get

Peeik · Oct 12, 2011

Hey just something very random......how do you guys write stuff like those equations so professionally on this website?? Please let me know so i can answer too!

AAEldar · Oct 12, 2011

Peeik said:
Hey just something very random......how do you guys write stuff like those equations so professionally on this website?? Please let me know so i can answer too!

Latex. When replying to a post click on the $f_x$ button and it'll open a new window.

Or go here:

http://www.codecogs.com/latex/eqneditor.php

Takes a little to get used to, but is so much better.

Finding the greatest term in binomial expansions HELP NEEDED! (2 Viewers)

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