p-x-1-x-1-2x-1-3x-1-15x-find-the-coefficient-of-the-x-3-term-in-the-expansion-

Question Number 179890 by mr W last updated on 03/Nov/22

p (x) = (1 - x) (1 + 2 x) (1 - 3 x) \dots . (1 - 15 x)

f i n d t h e c o e f f i c i e n t o f t h e x^{3} t e r m

i n t h e e x p a n s i o n .

Commented by mr W last updated on 03/Nov/22

a d a p t e d f r o m Q 179777

Answered by mr W last updated on 04/Nov/22

M e t h o d I I

w e n e e d t h e s e v a l u e s l a t e r :

\underset{k = 1}{\sum^{15}} {(- 1)}^{k} k = - (1 + 3 + 5 + \dots + 15) + (2 + 4 + \dots + 14)

= - \frac{8 \times 16}{2} + \frac{7 \times 16}{2} = - 8

\underset{k = 1}{\sum^{15}} k^{2} = 1^{2} + 2^{2} + \dots + 15^{2}

= \frac{15 \times (15 + 1) (2 \times 15 + 1)}{6} = 1240

\underset{k = 1}{\sum^{15}} {(- 1)}^{k} k^{3} = \underset{m = 0}{\sum^{7}} [- {(2 m + 1)}^{3} + {(2 m)}^{3}]

= \underset{m = 0}{\sum^{7}} [- 3 {(2 m)}^{2} - 3 (2 m) - 1]

= - \underset{m = 0}{\sum^{7}} [12 m^{2} + 6 m + 1]

= - [12 \times \frac{7 \times (7 + 1) \times (2 \times 7 + 1)}{6} + 6 \times \frac{7 \times 8}{2} + 8]

= - 1856

a s w e h a v e c a l c u l a t e d i n a n o t h e r

q u e s t i o n t h e c o e f . o f x^{2} i n p (x) i s

- 588 .

t h e c o e f . o f x^{3} i n p (x) i s

\frac{1}{3} \underset{k = 1}{\sum^{15}} {(- 1)}^{k} k \times [c o e f . o f x^{2} i n \frac{p (x)}{1 + {(- 1)}^{k} k x}]

= \frac{1}{3} \underset{k = 1}{\sum^{15}} {(- 1)}^{k} k \times [c o e f . o f x^{2} i n {p (x) - {(- 1)}^{k} k x p (x) + {(- 1)}^{2 k} k^{2} x^{2} p (x) - \dots}]

= \frac{1}{3} \underset{k = 1}{\sum^{15}} {(- 1)}^{k} k \times [- 588 - {(- 1)}^{k} k \underset{r = 1}{\sum^{15}} {(- 1)}^{r} r + k^{2}]

= \frac{1}{3} \underset{k = 1}{\sum^{15}} {(- 1)}^{k} k \times [- 588 + 8 {(- 1)}^{k} k + k^{2}]

= \frac{1}{3} \underset{k = 1}{\sum^{15}} [- 588 {(- 1)}^{k} k + 8 k^{2} + {(- 1)}^{k} k^{3}]

= \frac{1}{3} [588 \times 8 + 8 \times 1240 - 1856]

= \frac{1}{3} [12768]

= 4256

Commented by mr W last updated on 04/Nov/22

v e r i f i c a t i o n o f t h e a n s w e r :

Commented by mr W last updated on 04/Nov/22

Answered by mr W last updated on 04/Nov/22

M e t h o d I

b a s i c a l l y w e j u s t c a l c u l a t e t h e s u m

o f a l l p o s s i b l e p r o d u c t s f r o m a, b, c

w i t h a, b, c = (- 1, 2, - 3, 4, \dots, 14, - 15)

a n d a \neq b \neq c .

l e t a = {(- 1)}^{p} p, b = {(- 1)}^{q} q, c = {(- 1)}^{r} r

Σ Σ Σ a b c

= \underset{p = 1}{\sum^{15}} \underset{q = p + 1}{\sum^{15}} \underset{r = q + 1}{\sum^{15}} {(- 1)}^{p + q + r} p q r

= 4256

Commented by mr W last updated on 04/Nov/22

Answered by Peace last updated on 05/Nov/22

p (x) = \underset{n = 0}{\sum^{15}} a_{n} x^{n}

p (x) = \underset{k = 0}{\sum^{15}} \frac{p^{(k)} (0)}{k!} x^{k} w e w a n t \frac{p^{3} (0)}{6}

l e t ϵ > 0 e n o u g h s m a l l s u c h p (x) > 0 \forall x \in [- ϵ, ϵ]

i s p o s s i b l e p \in R_{15} [X] C o n t i n u s a n d p (0) = 1 > 0

f (x) = l n (p), = Σ l n (1 + k {(- 1)}^{k} x)

f^{'} = \frac{p^{'}}{p} = Σ \frac{k {(- 1)}^{k}}{1 + k {(- 1)}^{k} x}

f^{″} = \frac{p^{″}}{p} - \frac{p^{' 2}}{p^{2}} = - Σ \frac{k^{2}}{{(1 + k {(- 1)}^{k} x)}^{2}}

f^{‴} = \frac{p^{‴}}{p} - \frac{p^{'} p^{″}}{p^{2}} - \frac{2 p^{″} p^{'} p^{2} - 2 p^{' 3} p}{p^{4}} = 2 Σ \frac{k^{3} {(- 1)}^{k}}{{(1 + k {(- 1)}^{k} x)}^{3}}

p (0) = 1

f^{'} (0) = Σ k {(- 1)}^{k} = - 8

f^{″} (0) = - Σ k^{2} = \frac{- n (n + 1) (2 n + 1)}{6} = - 40.31 = - 1240

f^{‴} (0) = 2 Σ {(- 1)}^{k} k^{3} = 2 . - 1856 = - 3712

\frac{p^{'} (0)}{p (0)} = - 8 \Rightarrow p^{'} (0) = - 8

\frac{p^{″} (0)}{p (0)} - \frac{p^{' 2} (0)}{p^{2} (0)} = - 1240

\Rightarrow p^{″} (0) = - 1240 + 64 = - 1176

(3) . . p^{‴} (0) - 8.1176 - (2.1176.8 + 1024) = - 3712

p^{^{‴}} (0) - 24.1176 - 1024 = - 3712

p^{‴} (0) = 25536

\frac{p^{(3)} (0)}{6} = 4256

Commented by mr W last updated on 05/Nov/22

t h a n k s a l o t s i r!

t h a n k s a l o t s i r!

Commented by mindispower last updated on 05/Nov/22

w i t h e P l e a s u r h a v e a n i c e d a y

w i t h e P l e a s u r h a v e a n i c e d a y