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True or False: Assume that k = 1, 2, 3, --- . Then
(2)(4)(6)(8) ---(2k - 4)(2k - 2)(2k) = (2k)(k!)

2006-07-09 20:16:45 · 7 answers · asked by Anonymous in Science & Mathematics Mathematics

7 answers

THAT statement is FALSE.

(2)(4)(6)(8) = 384
(2)(4)(4!) = 192

However if the right side of the equation was (2^k)(k!) then it would be right.
(2^4)(4!) = 384

(2)(4)(6)(8).... = (2)(1)(2)(2)(2)(3)(2)(4) .... = 2^k (1)(2)(3)(4) ... = (2^k)(k!)

Using math induction we can also show that the new equation is true.

If using k=1 the formula is true and ....(2k-4)(2k-2)(2k)(2k+2) = 2^(k+1)((k+1)!)
then for all k >= 1 ....(2k-4)(2k-2)(2k) = (2^k)(k!) must be true.

To show k=1 is true ... 2 = (2^1)(1!) -> 2=2 TRUE
To show ....(2k-4)(2k-2)(2k)(2k+2) = 2^(k+1)((k+1)!)

First take the original equation
....(2k-4)(2k-2)(2k) = (2^k)(k!)
Multiply by 2k+2 on each side
....(2k-4)(2k-2)(2k)(2k+2) = (2^k)(k!)(2k+2)
Now if we can make the left side into (2^(k+1))((k+1)!) then we've shown that the orginal equation is true.
.... = (2^k)(k!)2(k+1)
.... = 2(2^k)(k!)(k+1)
.... = (2^(k+1))(k!)(k+1)
.... = (2^(k+1))((k+1)!) YAH, OUR NEW EQUATION IS TRUE!

2006-07-09 20:57:26 · answer #1 · answered by Michael M 6 · 0 0

True. Each factor in the expression (2)(4(6)(8) etc. is twice the size of the factors in a factorial (1)(2)(3)(4) etc. The number of factors is k, so we have k times 2 times the factorial, or (2k)(k1).

2006-07-09 20:25:44 · answer #2 · answered by Anonymous · 0 0

(2)(4)(6)(8)--- is part of (2k)(k!)
and so is (2k-4)(2k-2)(2k), because those are just the three numbers preceding 2k*k!.
Ex: (2k-2)=(2(k-1))
It's just a fancy way of writting it..., the (2k-4)(2k-2)(2k) would be included in the equation, regardless of wether or not they were stated, because if k=100, then they would be (196)(198)(200), and (2)(4)(6)(8)--- (196)(198)(200)=(2*100)(100!)

2006-07-09 21:03:11 · answer #3 · answered by Anonymous · 0 0

following michael m's reply

it should be (2^k)(k!)

if u factor a 2 out of every term of the LHS, u get:

(1*2)(2*2)(3*2)...{2(k-2)}{2(k-1)}(2k)
=(2*2*2*2*2...*2)(1*2*3*4*5...(k-2)(k-1)k)
=(2^k)(k!) by the definition of factorial

2006-07-10 01:27:01 · answer #4 · answered by angyansheng65537 2 · 0 0

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2016-11-06 03:19:54 · answer #5 · answered by ? 3 · 0 0

that statement is never correct
probably on R.H.S side 2k must be replaced by 2 power k;

2006-07-09 22:03:01 · answer #6 · answered by ? 2 · 0 0

true...lol

2006-07-09 20:20:14 · answer #7 · answered by Anonymous · 0 0

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