Binomial theorem question (1 Viewer)

Drsoccerball · Sep 8, 2015

rand_althor said:
$5(i) In the expansion of $(2x+5)^{20}$ in ascending powers of $x$ show that there are two successive terms with the same coefficient, which is the greatest of any of the coefficients. Express this coefficient in index form.$ \\ \\$I've found the coefficient to be $\binom{20}{15}2^55^{15}$. and found one of the terms to be $u_7$. To find the other do I just guess and check?.$

Just do the same process for the other value. Make sure you consider that x may=0 and therefore your answer would be wrong.

kawaiipotato · Sep 8, 2015

If I interpreted your question correctly, then you could do this.
$\frac{T_{r+1}}{T_r} > 1 $ for greatest coefficient $$

$$ Simplifying, we'll get $ r<15$

$$ So for $ T_{r+1} > T_r , r<15$

$$ For $ r = 14, T_{15} > T_{14}$

$r = 13, T_{14} > T_{13}$

$...$

$r = 1, T_{2} > T_{1}$

$$ So $ T_{15} > T_{14} > ... > T_{1}$

$To show there's a second one with the same coeff.$

$For \frac{T_{r+1}}{T_r} < 1, $ then $ r>15$

$$ So, $ T_{r+1} < T_r $ for $ r>15$

$r = 16, T_{17}< T_{16}$

$r = 17 T_{18} < T_{17}$

$...$

$\therefore T_{16} > ... > T_{20}$

$Since $ T_{16}$ is greater than one side of the polynomial and $ T_{15} $ is greater than the other side and they are consecutive, they must equal$

rand_althor · Sep 8, 2015

Drsoccerball said:
Just do the same process for the other value. Make sure you consider that x may=0 and therefore your answer would be wrong.

What do you mean by other value?

kawaiipotato said:
If I interpreted your question correctly, then you could do this.

Thanks!

rand_althor · Sep 8, 2015

Could someone also help with this?

$$If the $r$th term in the expansion of $(1+x)^n$ where $n$ is a positive integer is the greatest in absolute magnitude, show that $r$ is the first integer greater than $\frac{x(n+1)}{x+1}$.$

InteGrand · Sep 8, 2015

rand_althor said:
Could someone also help with this?

$$If the $r$th term in the expansion of $(1+x)^n$ where $n$ is a positive integer is the greatest in absolute magnitude, show that $r$ is the first integer greater than $\frac{x(n+1)}{x+1}$.$

First of all, do we start at r = 0, or at r = 1? Also, shouldn't that expression at the end have absolute value signs around it, since it could be negative?

rand_althor · Sep 8, 2015

InteGrand said:
First of all, do we start at r = 0, or at r = 1? Also, shouldn't that expression at the end have absolute value signs around it, since it could be negative?

I think you start at r=0. Not sure about the absolute value signs though. The textbook only says what I've posted.

InteGrand · Sep 8, 2015

$If we start from $r=0$, (rather than $r=1$), we have $T_r = \binom{n}{r}x^{r}$, $r=0,1,...,n$. Then $|T_r| = \left | \binom{n}{r}x^{r} \right |=\frac{n!}{r!(n-r)!} \left | x \right |^{r}$.$

$The term with largest absolute values satisfies the following simultaneously (assuming it is not $T_0$ or $T_{n}$; if $x=0$ though, $T_0$ would clearly be the largest term, and there may be other special cases where $T_0$ or $T_n$ would be the largest term. The formula suggested in fact does not work if $x=0$.):$

$$\frac{|T_r|}{|T_{r-1}|}\geq 1$\ \ (1)$

$and$

$\frac{|T_r|}{|T_{r+1}|}\geq 1.$\ \ (2)$

$We solve inequation (1) for $r$:$

$\frac{|T_r|}{|T_{r-1}|}\geq 1\Longleftrightarrow \frac{\frac{n!}{r!(n-r)!} \left | x \right |^{r}}{\frac{n!}{(r-1)!(n-r+1)!} \left | x \right |^{r-1}}>1$, $|x|\neq 0$

$$\Longleftrightarrow \frac{n-r+1}{r}|x|\geq 1$ (simplifying)$

$$r\leq \frac{|x|(n+1)}{1+|x|}$ (solving for $r$)$

$Now, we solve inequation (2) for $r$:$

$\frac{|T_r|}{|T_{r+1}|}\geq 1\Longleftrightarrow \frac{\frac{n!}{(r-1)!(n-r+1)!} \left | x \right |^{r-1}}{\frac{n!}{r!(n-r)!} \left | x \right |^{r}}\geq 1$, $|x|\neq 0$

$$\Longleftrightarrow \frac{r+1}{(n-r) \left | x \right |}\geq 1$ (simplifying)$

$$\Longleftrightarrow r\leq \frac{n|x|-1}{1+|x|}$ (solving for $r$).$

$$So we have $\frac{n|x|-1}{1+|x|}\leq r \leq \frac{|x|(n+1)}{1+|x|}$.$

$Notice that the difference between the right-hand bound and the left-hand bound is: $ \frac{|x|(n+1)}{1+|x|}-\frac{n|x|-1}{1+|x|} = \frac{n|x|+|x|-n|x|+1}{1+|x|}=\frac{1+|x|}{1+|x|}=1$. So we have $\frac{|x|(n+1)}{1+|x|}-1 \leq r \leq \frac{|x|(n+1)}{1+|x|}$.$

$Therefore, if the left-hand bound happens to be an integer, so is the right-hand bound, and vice versa. Note that the right-hand bound is always greater than 1, so this value of $r$ is bounded by two bounds which are positive and apart by distance 1. So $r$ here (the $r$ value of the greatest magnitude term) is a uniquely specified integer if the bounds are not integers, since there is one and only one integer between two non-integer values separated by 1. If the bounds \textit{are} integers, then $r$ may be either of the bound values, and we have two coefficients of equal magnitude. Either way, we see if we take $r$ to be the first integer greater than or equal to the lower bound, that is, $r=\left \lceil \frac{|x|(n+1)}{1+|x|}-1 \right \rceil$, or if we take $r=\left \lfloor \frac{|x|(n+1)}{1+|x|} \right \rfloor$ (the greatest integer less than or equal to the upper bound), we will obtain a coefficient with maximal magnitude.$

rand_althor · Sep 8, 2015

Thanks a lot! Could you also help with this one?

$$Prove that if $n$ is an even positive integer the coefficient of the $k$th term in the expansion of $(x+1)^n$ in descending powers of $x$ is the greatest when $k=\frac{n}{2}+1$. \\$

$$So far I have: $T_{k+1}=\binom{n}{k}x^{n-k}$, $T_{k}=\binom{n}{k-1}x^{n-k+1}$. Considering the coefficients: $\frac{T_{k+1}}{T_k}=\frac{n-k+1}{k}$. For the greatest coefficient $\frac{T_{k+1}}{T_k}>1\to\frac{n-k+1}{k}>1\to k <\frac{n}{2}+\frac{1}{2}$. Since $n$ is an even positive integer, it is divisible by 2, and hence $\frac{n}{2}$ is also an integer. So the greatest integer that satisifies the inequality would be $k=\frac{n}{2}$. I've made a mistake somewhere, but I can't figure out where.$

InteGrand · Sep 8, 2015

rand_althor said:
Thanks a lot! Could you also help with this one?

$$Prove that if $n$ is an even positive integer the coefficient of the $k$th term in the expansion of $(x+1)^n$ in descending powers of $x$ is the greatest when $k=\frac{n}{2}+1$. \\$

$$So far I have: $T_{k+1}=\binom{n}{k}x^{n-k}$, $T_{k}=\binom{n}{k-1}x^{n-k+1}$. Considering the coefficients: $\frac{T_{k+1}}{T_k}=\frac{n-k+1}{k}$. For the greatest coefficient $\frac{T_{k+1}}{T_k}>1\to\frac{n-k+1}{k}>1\to k <\frac{n}{2}+\frac{1}{2}$. Since $n$ is an even positive integer, it is divisible by 2, and hence $\frac{n}{2}$ is also an integer. So the greatest integer that satisifies the inequality would be $k=\frac{n}{2}$. I've made a mistake somewhere, but I can't figure out where.$

$Only mistake you made is that you used $T_{k+1}$ (rather than $T_k$) as your numerator, and hence greatest term, and so you should use $k+1$ as the term with the largest value (but you accidentally used $k$ instead, which is the mistake). So the largest term is the term $T_{k+1}=T_{\frac{n}{2}+1}$. (This is why I prefer to put $T_k$ on the numerator as the greatest term, so that, in general, when I find $k$, that's the answer, and no need to adjust it by adding 1 or anything.)$

rand_althor · Sep 8, 2015

InteGrand said:
$Only mistake you made is that you used $T_{k+1}$ (rather than $T_k$) as your numerator, and hence greatest term, and so you should use $k+1$ as the term with the largest value (but you accidentally used $k$ instead, which is the mistake). So the largest term is the term $T_{k+1}=T_{\frac{n}{2}+1}$. (This is why I prefer to put $T_k$ on the numerator as the greatest term, so that, in general, when I find $k$, that's the answer, and no need to adjust it by adding 1 or anything.)$

$$Thanks! So just to confirm: if I use $T_k=\binom{n}{k}x^{n-k}$, where $k=0,1,2...$ for $0\leq k \leq n$, and $T_{k-1}=\binom{n}{k-1}x^{n-(k-1)}$, where $k=1,2,3...$ for $0\leq k-1 \leq n$, the greatest coefficient can be determined using $\frac{T_k}{T_{k-1}}$? And this gives me a value of $k$ which, for $k=r$, is the $r$th term?$

Drsoccerball · Sep 8, 2015

Integrand for this particular question type can we assume x>1 since the so called "Greatest coefficient" isn't the greatest if x=0.00000000000000000000000000000000000000000001

InteGrand · Sep 8, 2015

Drsoccerball said:
Integrand for this particular question type can we assume x>1 since the so called "Greatest coefficient" isn't the greatest if x=0.00000000000000000000000000000000000000000001

This one asked for coefficient, so we're only looking at binomial coefficients.

And of course, we can also see that the statement we're asked to prove is true by looking at rows of Pascal's Triangle.

Binomial theorem question (1 Viewer)

rand_althor

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