## Re: [PrimeNumbers] solve using induction to find the sequence

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• In a message dated 29/04/03 22:51:04 GMT Daylight Time, luken2u@aol.com ... Assume that, for 0
Message 1 of 2 , Apr 29, 2003
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In a message dated 29/04/03 22:51:04 GMT Daylight Time, luken2u@...
writes:

> i have a problem that i can't figure out.. and i need to know how to
> do this b4 10 tonight... here is the problem (using induction)
>
> prove by induction
> let e (sub 0) = 1,
> e (sub 1) = 4
> and for n >1.
> let e sub n = 4[e (sub n - 1) - e (sub n-2)]
>
> what are the first six terms of the sequence?
>
> prove that e(sub n) = 2^n(n+1)
>

Assume that, for 0 <= k < n
e_k = (k+1)*2^k (1)

Then
e_n = 4*[e_(n-1) - e(n-2)]
= 4*[n*2^(n-1) - (n-1)*2^(n-2)]
= 2*n*2^n - (n-1)*2^n
= (n+1)*2^n
So if (1) is true for k < n, it is true also for k = n.
But (1) is true for n = 0 and n = 1.
So (1) is true for all n >= 0.
Q.E.D.

Mike Oakes

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