Answer: In Exercises 14, verify the formula

Chapter 9, Problem 4

(choose chapter or problem)

In Exercises 1-4, verify the formula.

\(\frac{1}{1\cdot3\cdot5\cdot\cdot(2k-5)}=\frac{2^kk!(2k-3)(2k-1)}{(2k)!},\quad\ \ k\ \ge\ 3\)

Text Transcription:

1 / 1 cdot 3 cdot 5 cdots (2k-5) = 2^k k! (2k-3)(2k-1) / (2k)!,  k geq 3

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