Prove-that-m-N-r-1-m-2m-2r-1-2r-1-2m-1-r-1-m-2m-2r-2r-2m-1-

Question Number 8013 by Yozzia last updated on 28/Sep/16

P r o v e t h a t, \forall m \in N,

\underset{r = 1}{\sum^{m}} (\begin{matrix} 2 m \\ 2 r - 1 \end{matrix}) {(2 r - 1)}^{2 m - 1} = \underset{r = 1}{\sum^{m}} (\begin{matrix} 2 m \\ 2 r \end{matrix}) {(2 r)}^{2 m - 1} .

Commented by FilupSmith last updated on 28/Sep/16

\frac{(2 m)! \times {(2 r - 1)}^{2 m - 1}}{(2 r - 1)! \times (2 x - 2 r + 1)!} = \frac{(2 m)! \times {(2 r)}^{2 m - 1}}{(2 r)! \times (2 x - 2 r)!}

\frac{(2 m)! \times {(2 r - 1)}^{2 m - 1}}{(2 r - 1)! \times (2 x - 2 r + 1)!} = \frac{(2 m)! \times {(2 r)}^{2 m - 2}}{(2 r - 1)! \times (2 x - 2 r)!}

\frac{{(2 r - 1)}^{2 m - 1}}{(2 x - 2 r + 1)!} = \frac{{(2 r)}^{2 m - 2}}{(2 x - 2 r)!}

\frac{{(2 r - 1)}^{2 m - 1}}{(2 x - 2 r + 1)!} = \frac{{(2 r)}^{2 m - 2}}{(2 x - 2 r)!}

\frac{{(2 r - 1)}^{2 m - 1}}{{(2 r)}^{2 m - 2}} = \frac{(2 x - 2 r + 1)!}{(2 x - 2 r)!}

\frac{{(2 r - 1)}^{2 m - 2} (2 r - 1)}{{(2 r)}^{2 m - 2}} = \frac{(2 x - 2 r + 1)!}{(2 x - 2 r)!}

{(\frac{2 r - 1}{2 r})}^{2 m - 2} (2 r - 1) = \frac{(2 x - 2 r + 1)!}{(2 x - 2 r)!}

i n c o n s t r u c t i o n

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

\underset{r = 1}{\sum^{m}} (\begin{matrix} 2 m \\ 2 r - 1 \end{matrix}) {(2 r - 1)}^{2 m - 1} = \underset{r = 1}{\sum^{m}} (\begin{matrix} 2 m \\ 2 r \end{matrix}) {(2 r)}^{2 m - 1} .

m = 2

^{4} C_{1} \cdot 1^{3} +^{4} C_{3} 3^{3} =^{4} C_{2} 2^{3} +^{4} C_{4} 4^{3}

4 \times 1 + 1 \times 27 \neq 6 \times 8 + 64 ?

Commented by Yozzia last updated on 30/Sep/16

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

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

O k . I made this error so many times .

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

{(1 - e^{x})}^{2 m} = \underset{k = 0}{\sum^{2 m}}^{2 m} C_{k} e^{k x} {(- 1)}^{k}

Let us write C_{k} f o r^{2 m} C_{k}

\frac{d^{2 m - 1}}{{d x}^{2 m - 1}} {(1 - e^{x})}^{2 m} = \underset{k = 1}{\sum^{2 m}}^{2 m} C_{k} k^{2 m - 1} {(- 1)}^{k}

index start from 1 since k = 0 is a constant

with derivative 0 .

LHS is 0 a t x = 0 since (1 - e^{x}) will be factor .

RHS

- C_{1} {(1)}^{2 m - 1} + C_{2} {(2)}^{2 m - 1} + - . . + C_{2 m} {(2 m)}^{2 m - 1} = 0

\Rightarrow \underset{r = 1}{\sum^{m}} C_{2 r} {(2 r)}^{2 m - 1} = \underset{r = 1}{\sum^{m}} C_{2 r - 1} {(2 r - 1)}^{2 m - 1}

odd terms equal to even term (excep C_{0})