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I have got a question which is stated as

Find numerically the greatest term in the expansion of $(2+3x)^9$ where $x=\frac{3}{2}$

I haven't any idea how to proceed but I just plugged in value of $x$ and got $$(2+3x)^9=\bigg(\frac{13}{2}\bigg)^9$$

How can I determine greatest term here?

Please help!!!

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    $\begingroup$ Hint: if $a_0 = \binom{9}{0}2^9 $ and $a_1 = \binom{9}{1}(9/4)^1 2^8 $, what do I need to multiply $a_0$ by to get $a_1$? What do I need to multiply $a_1$ by to get $a_2$? When is the number that I need to multiply by less than unity? $\endgroup$ – John Joy Jul 8 '17 at 11:57
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The term involving $(3/2)^n$ is

$$ a_n = {9 \choose n} 2^{9-n} \left( {9 \over 2} \right)^n$$

and so you have

$$ {a_n \over a_{n-1}} = {{9 \choose n} \over {9 \choose n-1}} {2^{9-n} \over 2^{9-(n-1)}} {(9/2)^n \over (9/2)^{n-1}} $$

or after some simplification

$$ {a_n \over a_{n-1}} = {{9 \choose n} \over {9 \choose n-1}} \times {9 \over 4}. $$

Now attack that quotient of binomial coefficients; it's

$$ {{9 \choose n} \over {9 \choose n-1}} = {9! \over n! (9-n)!} {(n-1)! (10-n)! \over 9!} = {(n-1)! \over n!} {(10-n)! \over (9-n)!} = {10 - n \over n}.$$

So you have

$$ {a_n \over a_{n-1}} = {10 - n \over n} \times {9 \over 4} $$

and this decreases as $n$ increases. So $a_n/a_{n-1} > 1$ for small $n$ and $a_n/a_{n-1} < 1$ for large $n$ - the sequence of $a_n$ increases and then decreases as $n$ increases. Find the $n$ where this changeover happens.

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You have $$(3 x+2)^9=19683 x^9+118098 x^8+314928 x^7+489888 x^6+489888 x^5+326592 x^4+145152 x^3+41472 x^2+6912 x+512$$ Plugging $x=1.5$ the terms, from degree $0$ up are $$512,10368,93312,489888,1.65337\times 10^6,3.72009\times 10^6,5.58013\times 10^6,5.38084\times 10^6,3.02672\times 10^6,756681$$ the largest terms is for $x^6$ and is $5.58013\times 10^6$

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  • $\begingroup$ Don't you think it is a bit lengthy way to do in exams , I would not have calculators there $\endgroup$ – Atul Mishra Jul 8 '17 at 11:31
  • $\begingroup$ My calculations give that the largest term is $5580130.5$, corresponding to $489888x^6$. $\endgroup$ – lhf Jul 8 '17 at 13:26
  • $\begingroup$ @AtulMishra to anwer a dumb question like this cheating is allowed :) $\endgroup$ – Raffaele Jul 8 '17 at 19:03
  • $\begingroup$ And what if I say I was not the only dumb in the examination hall @Raffaele? $\endgroup$ – Atul Mishra Jul 9 '17 at 2:20
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HINT:

The middle term of the expansion is the greatest. For odd $n$, the greatest term will be for $r=\frac{n+1}{2}$ and for even $n$, there will be two greatest terms and they will be for $r=\frac{n}{2}$ and $r=\frac{n}{2}+1$.

Hope this helps.

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  • $\begingroup$ Yes I can find the solution from here $\endgroup$ – Atul Mishra Jul 8 '17 at 11:05
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    $\begingroup$ The middle coefficient of the expansion is the greatest. It is not clear that when you add the powers of $2$, $3$ and $x$, this will still be true. In fact, it is not: the greatest term corresponds to $x^6$ when $x=3/2$. $\endgroup$ – lhf Jul 8 '17 at 11:06

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