Recurrence relation

[valid mica]

Why is the bit highlighted true. I can’t derive why it holds

[errant zenith]

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[torn glen]

2^n n! = prod (1<=i<=n) 2 x prod (1<=i<=n) i = prod (1<=i<=n) (2i) = 2n x (2n-2) x... x 2

[torn sparrow]

because you can factor out 2 to get $\prod_{1\leq k\leq n} 2(n-k)$

[leaden atlas]

;( | 追放された興奮

[torn sparrow]

therefore you get $\prod_{k=1}^{n} 2 \cdot \prod_{k=1}^{n} n-k=2^n\cdot n!$

[leaden atlas]

;( | 追放された興奮

[silk junco]

this is 0

$2^n(n!)=(2\times n)(2\times(n-1))\ldots(2\times 2)(2\times 1)$ as stated

[leaden atlas]

confusedffey

[silk junco]

we multiply a 2 to every term in the factorial in a sense

[torn sparrow]

$\prod_{k=0}^{n} 2 \cdot \prod_{k=0}^{n-1} n-k=2^n\cdot n!$

[leaden atlas]

;( | 追放された興奮

[silk junco]

you have 2^(n+1) there

[torn sparrow]

bro

$\prod{k=0}^{n-1} 2 \cdot \prod{k=0}^{n-1} n-k=2^n\cdot n!$

[leaden atlas]

;( | 追放された興奮

[silk junco]

@torn sparrow don't convolute help posts with erroneous things over and over, it'll confuse op for no reason.

use #bot-cmd-latex to test timings out

[torn sparrow]

wait whats wrong with this one?

oh.

[torn glen]

I think that OP left for new adventures, never to return.

[mystic epoch]

@valid mica

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[valid mica]

+close

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[silk junco]

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