what-is-the-largest-coefficient-of-4-3x-5-

Question Number 125697 by aurpeyz last updated on 13/Dec/20

w h a t i s t h e l a r g e s t c o e f f i c i e n t o f

{(4 + 3 x)}^{- 5} ?

Answered by mr W last updated on 13/Dec/20

{(4 + 3 x)}^{- 5}

= 4^{- 5} {[1 - (- \frac{3 x}{4})]}^{- 5}

= \frac{1}{1024} \underset{k = 0}{\sum^{\infty}} C_{4}^{k + 4} {(- \frac{3 x}{4})}^{k}

= \frac{1}{1024} \underset{k = 0}{\sum^{\infty}} {(- 1)}^{k} C_{4}^{k + 4} {(\frac{3}{4})}^{k} x^{k}

= \underset{k = 0}{\sum^{\infty}} a_{k} x^{k}

a_{k} = \frac{{(- 1)}^{k}}{1024} C_{4}^{k + 4} {(\frac{3}{4})}^{k}

p o s i t i v e c o e f . w h e n k = e v e n = 2 n

n e g a t i v e c o e f . w h e n k = o d d = 2 n + 1

t o f i n d t h e l a r g e s t c o e f . w e o n l y

n e e d t o l o o k a t p o s i t i v e c o e f f i c i e n t s,

i . e . k = 2 n .

a_{2 n} = \frac{1}{1024} C_{4}^{2 n + 4} {(\frac{3}{4})}^{2 n}

a_{2 (n + 1)} = \frac{1}{1024} C_{4}^{2 n + 6} {(\frac{3}{4})}^{2 (n + 1)}

t o f i n d t h e l a r g e s t c o e f . w e n e e d t o

f i n d t h e s m a l l e s t n s a t i s f y i n g

a_{2 n} > a_{2 (n + 1)}

i . e .

\frac{1}{1024} C_{4}^{2 n + 4} {(\frac{3}{4})}^{2 n} > \frac{1}{1024} C_{4}^{2 n + 6} {(\frac{3}{4})}^{2 (n + 1)}

C_{4}^{2 n + 4} > C_{4}^{2 n + 6} {(\frac{3}{4})}^{2}

\frac{(2 n + 4)!}{4! (2 n)!} > \frac{(2 n + 6)!}{4! (2 n + 2)!} {(\frac{3}{4})}^{2}

\frac{16}{9} > \frac{(n + 3) (2 n + 5)}{(n + 1) (2 n + 1)}

16 (2 n^{2} + 3 n + 1) > 9 (2 n^{2} + 11 n + 15)

14 n^{2} - 51 n - 119 > 0

n > \frac{51 + \sqrt{51^{2} + 4 \times 14 \times 119}}{28} \approx 5.3

t h e s m a l l e s t n i s 6, i . e . t h e l a r g e s t

c o e f f i c i e n t i s a_{12} :

a_{12} = \frac{1}{1024} C_{4}^{16} {(\frac{3}{4})}^{12} = \frac{241 805 655}{4 294 967 296}

y o u c a n c h e c k t h i s w i t h W o l f r a m A l p h a :

Commented by mr W last updated on 13/Dec/20

Commented by aurpeyz last updated on 13/Dec/20

w o w . t h i s i s s o e x p l i c i t . i c a n t

a p p r e c i a t e y o u e n o u g h . t h a n k s

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