- #1

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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 .

- Thread starter Mithal
- Start date

- #1

- 28

- 0

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 .

- #2

Hurkyl

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[tex]

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

= \frac{4^n - 1}{2n + 1}

[/tex]

?

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

- #3

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