Prove-that-i-0-n-1-n-1-i-i-n-1-i-1-0-for-n-N-

Question Number 7861 by Yozzia last updated on 21/Sep/16

P r o v e t h a t \underset{i = 0}{\sum^{n + 1}} (\begin{matrix} n + 1 \\ i \end{matrix}) i^{n} {(- 1)}^{i + 1} = 0

f o r \forall n \in N .

Commented by FilupSmith last updated on 22/Sep/16

a t t e m p t

\underset{i = 0}{\sum^{n + 1}} (\begin{matrix} n + 1 \\ i \end{matrix}) = \underset{i = 0}{\sum^{k}} \frac{k!}{i! \cdot (k - i)!} = k! \underset{i = 0}{\sum^{k}} \frac{1}{i! \cdot (k - i)!}

n + 1 = k \Rightarrow n = k - 1

\underset{i = 0}{\sum^{k}} S = k! (\frac{1}{0! \times k!} + \frac{1}{1! \times (k - 1)!} + \frac{1}{2! \times (k - 2)!} + \dots + \frac{1}{(k - 1)! \times 1!} + \frac{1}{k! \times 0!})

∴ \underset{i = 0}{\sum^{n + 1}} (\begin{matrix} n + 1 \\ i \end{matrix}) i^{n} {(- 1)}^{n + 1} = \underset{i = 0}{\sum^{k}} {S i}^{k - 1} {(- 1)}^{k}

\underset{i = 0}{\sum^{k}} {S i}^{k - 1} = k! (\frac{0^{k - 1}}{0! \times k!} + \frac{1^{k - 1}}{1! \times (k - 1)!} + \frac{2^{k - 1}}{2! \times (k - 2)!} + \dots + \frac{{(k - 1)}^{k - 1}}{(k - 1)! \times 1!} + \frac{k^{k - 1}}{k! \times 0!})

\underset{i = 0}{\sum^{k}} {S i}^{k - 1} = \frac{0^{k - 1}}{0!} + \frac{k \times 1^{k - 1}}{1!} + \frac{k (k - 1) 2^{k - 1}}{2!} + \frac{k (k - 1) (k - 2) 3^{k - 1}}{3!} + \dots + \frac{k {(k - 1)}^{k - 1}}{1!} + \frac{k^{k - 1}}{0!}

∴ \underset{i = 0}{\sum^{k}} {S i}^{k - 1} {(- 1)}^{k} = \frac{0^{k - 1}}{0!} + \frac{k \times 1^{k - 1}}{1!} - \frac{k (k - 1) 2^{k - 1}}{2!} + \frac{k (k - 1) (k - 2) 3^{k - 1}}{3!} - \dots + \frac{k {(k - 1)}^{k - 1}}{1!} - \frac{k^{k - 1}}{0!}

note : + if previous denominator has a even term

∴ = \frac{0^{k - 1}}{0!} + (\frac{k \times 1^{k - 1}}{1!} + \frac{k (k - 1) (k - 2) 3^{k - 1}}{3!} + \dots) - (\frac{k (k - 1) 2^{k - 1}}{2!} + \frac{k (k - 1) (k - 2) (k - 3) 4^{k - 1}}{4!} + \dots)

w o r k i n g

c u r r e n t l y u n s u r e h o w t o c o n t i n u e

Commented by prakash jain last updated on 01/Oct/16

f (x) = {(1 - e^{x})}^{n + 1}

= - \underset{i = 0}{\sum^{n + 1}}^{n + 1} C_{i} e^{i x} {(- 1)}^{i + 1}

f^{n} (x) = - \underset{i = 0}{\sum^{n}}^{n + 1} C_{i} i^{n} e^{i x} {(- 1)}^{i + 1} = required expression

at x = 0

Also f (x) = {(1 - e^{x})}^{n + 1} = L H S

f^{'} (x) = - (n + 1) {(1 - e^{x})}^{n} (- e^{x})

f^{2} (x) = - [(n + 1) (e^{x}) {(1 - e^{x})}^{n} - (n + 1) {n e}^{x} {(1 - e^{x})}^{n - 1} (- e^{x})]

⋮

f^{n} (0) = 0 [∵ (1 - e^{x}) is factor till n^{t h} d i f f e r e n t i a t i o n]

hence

- \underset{i = 0}{\sum^{n}}^{n + 1} C_{i} i^{n} {(- 1)}^{i + 1}

please check

Commented by Yozzia last updated on 30/Sep/16

T h a t s e e m s s u f f i c i e n t t o s h o w t h a t

t h e e q u a t i o n i s t r u e . N i c e l y d o n e!

I l i k e t h e a p p l i c a t i o n o f t h e e x p o n e n t i a l

f u n c t i o n a n d b i n o m i a l t h e o r e m .

Answered by prakash jain last updated on 22/Sep/16

n = 2

-^{3} C_{0} . 0 + 1^{2}^{3} C_{1} - 2^{2}^{3} C_{2} + 3^{2}^{3} C_{3}

= - 0 + 3 - 12 + 9 = 0

n = 3

- 0 + 1^{3} \times^{4} C_{1} - 2^{3} \times^{4} C_{2} + 3^{3} \times^{4} C_{3} - 4^{3} \times^{4} C_{4}

= 1 \times 4 - 8 \times 6 + 27 \times 4 - 64

= 4 - 48 - 108 - 64 = 0

Commented by Yozzia last updated on 22/Sep/16

(\begin{matrix} 3 \\ 0 \end{matrix}) h a s c o e f f i c i e n t 0 f o r T (n) = (\begin{matrix} n + 1 \\ i \end{matrix}) i^{n} {(- 1)}^{n + 1} n ⩾ 0 .

Commented by Yozzia last updated on 22/Sep/16

I m e t t h i s s e r i e s i n p r o v i n g b y i n d u c t i o n t h a t

t h e g e n e r a l c o e f f i c i e n t a_{n} o f t h e t e r m s i n t h e p o w e r s e r i e s

o f t h e W - L a m b e r t f u n c t i o n i s

a_{n} = \frac{{(- 1)}^{n + 1} n^{n - 2}}{(n - 1)!}, n ⩾ 1

W (x) = \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + 1} n^{n - 2} x^{n}}{(n - 1)!} {- 1 ⩽ x ⩽ 1}

W (x) = f^{- 1} ({x e}^{x}) f o r - 1 ⩽ x ⩽ 1 .

Commented by prakash jain last updated on 22/Sep/16

Ok my mistake will look at again .

Commented by FilupSmith last updated on 22/Sep/16

This is too complicated . I {don}^{'} t know

much about the W - F u n c t i o n

Commented by prakash jain last updated on 22/Sep/16

n = 3 n o t z e r o .

Commented by Yozzia last updated on 22/Sep/16

C h e c k t h a t 3^{3} t e r m .

Commented by prakash jain last updated on 30/Sep/16

Please check answer in comment .