Find-the-greatest-coefficient-in-the-following-without-actually-expand-i-5-3x-10-ii-5-3x-10-

Question Number 102278 by I want to learn more last updated on 08/Jul/20

Find the greatest coefficient in the following without

actually expand .

(i) {(5 - 3 x)}^{10}

(ii) {(5 + 3 x)}^{- 10}

Commented by mr W last updated on 08/Jul/20

s e e Q 94397

Commented by mr W last updated on 08/Jul/20

s e e Q 91464

Commented by I want to learn more last updated on 08/Jul/20

Sir am going say 5^{10} \underset{k = 0}{\sum^{10}} \overset{10}{} C_{k} {(- \frac{3 x}{5})}^{k}

Sir am going say 5^{10} \underset{k = 0}{\sum^{10}} \overset{10}{} C_{k} {(- \frac{3}{5})}^{k} . x^{k}

Am concern about the minus .

And here,

{(3 + 2 x)}^{- 7}

Sir am going say \frac{1}{3^{7}} \underset{k = 0}{\sum^{\infty}} \overset{k + 6}{} C_{k} {(- \frac{2 x}{3})}^{k}

The signs is my concern .

Just like the opposite sign to the one you solved sir .

Commented by mr W last updated on 08/Jul/20

i s h o w e d y o u t w o d i f f e r e n t w a y s h o w

t o t r e a t t h e “ -^{″} s i g n .

Answered by john santu last updated on 08/Jul/20

(1) C_{5}^{10} 5^{5} {(- 3 x)}^{6} = \frac{10.9.8.6}{5.4.3.2.1} . 5^{5} . {(- 3)}^{6}

= 5^{5} . 72 . 3^{6}

Commented by I want to learn more last updated on 08/Jul/20

Thanks sir

Answered by mr W last updated on 08/Jul/20

(i)

{(5 - 3 x)}^{10} = \underset{k = 0}{\sum^{10}} C_{k}^{10} 5^{10 - k} {(- 3 x)}^{k}

= \underset{k = 0}{\sum^{10}} {(- 1)}^{k} C_{k}^{10} 3^{k} 5^{10 - k} x^{k}

a_{k} = {(- 1)}^{k} C_{k}^{10} 3^{k} 5^{10 - k}

f o r g r e a t e s t c o e f f i c i e n t a_{k} w e s h o u l d

o n l y l o o k a t t h e p o s i t i v e v a l u e s, i . e .

i f k i s e v e n : k = 2 n

a_{2 n} = C_{2 n}^{10} 3^{2 n} 5^{10 - 2 n}

= 5^{10} C_{2 n}^{10} {(\frac{9}{25})}^{n}

l e t C_{2 n}^{10} {(\frac{9}{25})}^{n} > C_{2 (n + 1)}^{10} {(\frac{9}{25})}^{n + 1}

\frac{10!}{(2 n)! (10 - 2 n)!} {(\frac{9}{25})}^{n} > \frac{10!}{(2 n + 2)! (10 - 2 n - 2)!} {(\frac{9}{25})}^{n + 1}

\frac{1}{(10 - 2 n) (10 - 2 n - 1)} > \frac{1}{(2 n + 2) (2 n + 1)} \times \frac{9}{25}

\frac{25}{45 - 19 n + 2 n^{2}} > \frac{9}{2 n^{2} + 3 n + 1}

16 n^{2} + 123 n - 190 > 0

n > \frac{- 123 + \sqrt{123^{2} + 4 \times 16 \times 190}}{32} \approx 1.3

\Rightarrow n = 2

i . e . a_{4} i s t h e g r e a t e s c o e f f i c i e n t .

a_{4} = 5^{10} \times C_{4}^{10} \times {(\frac{9}{25})}^{2} = 265 781 250

Commented by I want to learn more last updated on 09/Jul/20

I really appreciate sir

Answered by mr W last updated on 08/Jul/20

(i i)

{(5 + 3 x)}^{- 10}

= \frac{1}{5^{10}} {(1 + \frac{3 x}{5})}^{- 10}

= \frac{1}{5^{10}} \underset{k = 0}{\sum^{\infty}} C_{9}^{k + 9} {(- \frac{3}{5})}^{k} x^{k}

l e t C_{9}^{k + 9} {(\frac{3}{5})}^{k} > C_{9}^{k + 10} {(\frac{3}{5})}^{k + 1}

\frac{(k + 9)!}{9! k!} > \frac{(k + 10)!}{9! (k + 1)!} \times \frac{3}{5}

1 > \frac{(k + 10)}{(k + 1)} \times \frac{3}{5}

2 k > 25

k > \frac{25}{2} = 12.5

\Rightarrow k = 14 s i n c e w e s e a r c h e v e n t e r m s

i . e . a_{14} i s t h e g r e a t e s t c o e f f i c i e n t .

a_{14} = \frac{1}{5^{10}} \times C_{9}^{23} \times {(\frac{3}{5})}^{14} = \frac{817190 \times 3^{14}}{5^{24}}

Commented by I want to learn more last updated on 09/Jul/20

I really appreciate sir