The succession [Xsubn] is defined as

Xsub0 = 0, Xsub(n+1) = Xsubn + n(n+1)

Show that Xsubn = (1/3)(n-1)n(n+1).

I am assuming I am not allowed to use what I am demonstring as part of the proof. No clue what do despite staring at it for 15 minutes.

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- Oct 12th 2009, 12:20 PMHeadOnAPikeDon't know what to do here, recursive problem
The succession [Xsubn] is defined as

Xsub0 = 0, Xsub(n+1) = Xsubn + n(n+1)

Show that Xsubn = (1/3)(n-1)n(n+1).

I am assuming I am not allowed to use what I am demonstring as part of the proof. No clue what do despite staring at it for 15 minutes. - Oct 12th 2009, 12:51 PMTaluivren
Hi,

,

and we have to prove .

We proceed by induction:

.. so the base case holds.

suppose the statement holds for and let's prove it holds for :

(by induction hypotheses)

so the inductive step is done. - Oct 12th 2009, 01:30 PMHeadOnAPike
My mistake was thinking you can't use the statement your trying to prove. But with induction you can. Thanks.

- Oct 12th 2009, 08:28 PMHallsofIvy
But it is perfectly valid to show that what you are given is a solution to an equation by putting it into the equation.

For example, it would be very difficult to**solve**but it is very easy to show that x= 2 is a solutions: [tex]2^2- 6(2^2)- 3(2)- 6= 32- 24- 6- 2= 0.

Similarly, here, you are not asked to solve the recursion, just to show that satisfies it. And that is basically what Taluivren did.