find-the-largest-and-smallest-coefficient-in-4-3x-5-

Question Number 114739 by mr W last updated on 20/Sep/20

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

c o e f f i c i e n t i n {(4 + 3 x)}^{- 5} .

Answered by mr W last updated on 20/Sep/20

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

= 4^{- 5} \underset{k = 0}{\sum^{\infty}} {(- 1)}^{k} C_{4}^{k + 4} {(\frac{3}{4})}^{k} x^{k}

w e s e e t h e e v e n t e r m s h a v e p o s i t i v e

c o e f f i c i e n t s a n d t h e o d d t e r m s h a v e

a l w a y s n e g a t i v e c o e f f i c i e n t s .

e v e n t e r m s : k = 2 n w i t h n = 0, 1, 2, 3, \dots

a_{2 n} = 4^{- 5} C_{4}^{2 n + 4} {(\frac{3}{4})}^{2 n}

l e t a_{2 n} > a_{2 (n + 1)}

4^{- 5} C_{4}^{2 n + 4} {(\frac{3}{4})}^{2 n} > 4^{- 5} C_{4}^{2 (n + 1) + 4} {(\frac{3}{4})}^{2 (n + 1)}

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

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

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

n > \frac{51 + \sqrt{51^{2} + 4 \times 15 \times 119}}{30} \approx 4.99

\Rightarrow n ⩾ 5, 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 f o r t h e t e r m x^{10} :

a_{10} = 4^{- 5} C_{4}^{14} {(\frac{3}{4})}^{10} = \frac{1001 \times 3^{10}}{4^{15}}

= \frac{59 108 049}{1 073 741 824}

o d d t e r m s : k = 2 n + 1

a_{2 n + 1} = - 4^{- 5} C_{4}^{2 n + 5} {(\frac{3}{4})}^{2 n + 1}

l e t a_{2 n + 1} < a_{2 (n + 1) + 1}

- 4^{- 5} C_{4}^{2 n + 5} {(\frac{3}{4})}^{2 n + 1} < - 4^{- 5} C_{4}^{2 n + 7} {(\frac{3}{4})}^{2 n + 3}

C_{4}^{2 n + 5} > C_{4}^{2 n + 7} {(\frac{3}{4})}^{2}

\frac{16}{9} > \frac{(n + 3) (2 n + 7)}{(n + 1) (2 n + 3)}

14 n^{2} - 37 n - 141 > 0

n > \frac{37 + \sqrt{37^{2} + 4 \times 14 \times 141}}{28} \approx 4.8

\Rightarrow n ⩾ 5, i . e . t h e s m a l l e s t c o e f f i c i e n t

i s f o r t h e t e r m x^{11} .

a_{11} = - 4^{- 5} C_{4}^{15} {(\frac{3}{4})}^{11} = - \frac{1365 \times 3^{11}}{4^{16}}

= - \frac{241 805 655}{4 294 967 396}

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