Can-We-expand-the-following-expression-1-x-1-2x-1-3x-1-nx-or-is-there-any-formula-for-this-

Question Number 30267 by Nayon.Sm last updated on 19/Feb/18

C a n W e e x p a n d t h e f o l l o w i n g

e x p r e s s i o n ?

(1 + x) (1 + 2 x) (1 + 3 x) \dots \dots (1 + n x)

o r i s t h e r e a n y f o r m u l a f o r t h i s ?

Commented by Penguin last updated on 19/Feb/18

f (x) = \underset{k = 1}{\prod^{n}} (1 + k x)

= (1 + x) (1 + 2 x) (1 + 3 x) \dots (1 + n x)

= x^{n} (1 + \frac{1}{x}) \dots (n + \frac{1}{x})

= x^{n} {(1 + \frac{1}{x})}_{n}

{(x)}_{n} \equiv \frac{Γ (x + n)}{Γ (x)} = x (x + 1) \dots (x + n - 1)

∴ f (x) = x^{n} Γ (\frac{1}{x} + n) Γ {(x)}^{- 1}

Commented by Nayon.Sm last updated on 19/Feb/18

i t s n o t

Commented by Penguin last updated on 19/Feb/18

Why ?

Commented by Nayon.Sm last updated on 19/Feb/18

w h a t i s Γ (x) \overset{}{?}

w h a t i s

Commented by Penguin last updated on 19/Feb/18

Gamma function

Γ (x) = (x - 1)! = (x - 1) \times (x - 2) \times \dots \times 3 \times 2 \times 1

Commented by abdo imad last updated on 20/Feb/18

Γ (x) i s t h e e u l e r f u n c t i o n d e f i n e d b y

Γ (x) = \int_{0}^{\infty} t^{x - 1} e^{- t} d t f o r x > 0 .

Commented by Nayon.Sm last updated on 10/Mar/18

Yes