Proof of Combinatorial Problem: Summation from k=1 to n

[Mithal]

I have this problem as follow

prove that

Summation of k=1 to n to the following term

( (-1)^(k+1) (( 2n-k) C ( k-1)) (4^(n-k))/k ) = ((4^n) - 1)/(2 n +1)

Note that the symbol C above meant the symbol of combination .

 

[Hurkyl]

I assume you mean:

$$\sum_{k = 1}^{n} (-1)^{k+1} \binom{2n-k}{k-1} \frac{1}{k} 4^{n-k} = \frac{4^n - 1}{2n + 1}$$

?

Have you tried anything? Or at least thought about how to begin, even if you weren't able to carry it through?

 

[Mithal]

Yes , I tried to do it using induction combined with Pascal's identity but it seems it doesn't work . Any suggestions how to go through ?