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I know that $2\cdot 4\cdot 6\cdot 8 \cdot \ldots \cdot (2k)$ is $2^kk!$ but what is the value of these terms up to the $(2k+2)^\text{th}$ term?

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    $\begingroup$ Just replace the $k$ in the expression $2^kk!$ with $k+1$. $\endgroup$
    – rotten
    Commented Sep 14, 2014 at 14:29
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    $\begingroup$ What is $2^kk!\times(2k+2)$? (write $2k+2=2(k+1)$) $\endgroup$ Commented Sep 14, 2014 at 14:31
  • $\begingroup$ @rotten: So if I want to calculate $2\cdot 4\cdot 6\cdot 8 \cdot \ldots \cdot (2k-2)$ I need to replace my $k$ with $k-1$? i.e $2\cdot 4\cdot 6\cdot 8 \cdot \ldots \cdot (2k-2)=2^{k-2}\cdot(k-2)!$? $\endgroup$
    – E Be
    Commented Sep 14, 2014 at 14:34
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    $\begingroup$ Yes, you do. I have to write more or it won't let me post, so, yeah. Yes, you do. $\endgroup$
    – rotten
    Commented Sep 14, 2014 at 14:35

1 Answer 1

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$$2\cdot 4\cdot 6\cdot 8 \cdot \ldots \cdot (2k+2)~=~(2\cdot\underline1)(2\cdot\underline2)(2\cdot\underline3)(2\cdot\underline4)\ldots\big(2\cdot[k+1]\big)~=~2^{k+1}(k+1)!$$

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