Find the greatest coefficient in the expansion of (6 − 4x)^(−3)


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Question Number 62945 by Tawa1 last updated on 27/Jun/19

$Find the greatest coefficient in the expansion of {(6 - 4 x)}^{- 3}$
	Answered by mr W last updated on 27/Jun/19

	${(6 - 4 x)}^{- 3}$ $= 6^{- 3} {(1 - \frac{2 x}{3})}^{- 3}$ $= 6^{- 3} \underset{n = 0}{\sum^{\infty}} C_{2}^{n + 2} {(\frac{2}{3})}^{n} x^{n}$ $= 6^{- 3} \underset{n = 0}{\sum^{\infty}} a_{n} x^{n}$ $a_{n} = C_{2}^{n + 2} {(\frac{2}{3})}^{n} = \frac{(n + 2) (n + 1)}{2} {(\frac{2}{3})}^{n}$ $a (x) = \frac{(x + 2) (x + 1)}{2} {(\frac{2}{3})}^{x}$ $a (x) h a s m a x i m u m a t x = 3.48$ $\Rightarrow a_{n} h a s m a x i m u m a t n = 3 o r 4$ $a_{3} = \frac{5 \times 4}{2} \times {(\frac{2}{3})}^{3} = \frac{80}{27}$ $a_{4} = \frac{6 \times 5}{2} \times {(\frac{2}{3})}^{4} = \frac{80}{27}$ $\Rightarrow x^{3} a n d x^{4} t e r m h a s t h e l a r g e s t c o e f .$ $w h i c h i s 6^{- 3} \times \frac{80}{27} = \frac{10}{729} .$
	Commented by Tawa1 last updated on 27/Jun/19

	$God bless you sir$
	Commented by Tawa1 last updated on 29/Jun/19

	$Sir, how did you get . 3.48 and 3 or 4$
	Commented by mr W last updated on 29/Jun/19

	$l o o k a t a (x) = \frac{(x + 2) (x + 1)}{2} {(\frac{2}{3})}^{x}$ $(x + 2) (x + 1) i s i n c r e a s i n g w i t h x,$ ${(\frac{2}{3})}^{x} i s d e c r e a s i n g w i t h x,$ $s o a n (x) h a s a m a x i m u m a t s o m e$ $p o i n t, a t t h i s p o i n t \frac{d a (x)}{d x} = 0 .$ $\frac{d a (x)}{d x} = \frac{(2 x + 3)}{2} {(\frac{2}{3})}^{x} + \frac{(x^{2} + 3 x + 1)}{2} {(\frac{2}{3})}^{x} \ln (\frac{2}{3}) = 0$ $(2 x + 3) + (x^{2} + 3 x + 1) \ln (\frac{2}{3}) = 0$ $\ln (\frac{2}{3}) x^{2} + [3 \ln (\frac{2}{3}) - 2] x + [\ln (\frac{2}{3}) - 3] = 0$ $\Rightarrow x = \frac{- 3 \ln (\frac{2}{3}) + 2 + \sqrt{{[3 \ln (\frac{2}{3}) - 2]}^{2} - 4 \ln (\frac{2}{3}) [\ln (\frac{2}{3}) - 3]}}{2 \ln (\frac{2}{3})} = 3.48$ $i . e . a (x) h a s m a x i m u m a t x = 3.48$ $a_{n} w i l l h a v e m a x i m u m w h e n n i s$ $s o c l o s e t o 3.48 a s p o s s i b l e, t h a t i s$ $n = 3 o r n = 4 .$
	Commented by Tawa1 last updated on 29/Jun/19

	$God bless you sir$
	Commented by mr W last updated on 13/Jul/19

	$a n o t h e r w a y t o f i n d t h e l a r g e s t c o e f . :$ $a_{n} = \frac{(n + 2) (n + 1)}{2} {(\frac{2}{3})}^{n}$ $a_{n - 1} = \frac{(n + 1) n}{2} {(\frac{2}{3})}^{n - 1}$ $a_{n + 1} = \frac{(n + 3) (n + 2)}{2} {(\frac{2}{3})}^{n + 1}$ $s u c h t h a t a_{n} i s m a x i m u m,$ $\frac{a_{n}}{a_{n - 1}} ⩾ 1 :$ $\frac{(n + 2) (n + 1)}{(n + 1) n} (\frac{2}{3}) ⩾ 1$ $\frac{n + 2}{n} ⩾ \frac{3}{2}$ $2 n + 4 ⩾ 3 n$ $\Rightarrow n ⩽ 4$ $\frac{a_{n}}{a_{n + 1}} ⩾ 1 :$ $\frac{(n + 2) (n + 1)}{(n + 3) (n + 2)} {(\frac{2}{3})}^{- 1} ⩾ 1$ $\frac{n + 1}{n + 3} ⩾ \frac{2}{3}$ $3 n + 3 ⩾ 2 n + 6$ $\Rightarrow n ⩾ 3$ $\Rightarrow n = 3 o r 4$
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