Any suggestions for an approach to proving that the nth term of the following sequence:

1,2,2,3,3,3,4,4,4,4,5,5,5...

is, where[ ((2*n)^(1/2) + (1/2)) ][x]denotes thefloor value of xand^denotesto the power of

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- Sep 9th 2008, 02:59 AMp vs npnth term of a sequence
Any suggestions for an approach to proving that the nth term of the following sequence:

1,2,2,3,3,3,4,4,4,4,5,5,5...

is, where*[ ((2*n)^(1/2) + (1/2)) ]**[x]*denotes the*floor value of x*and*^*denotes*to the power of* - Sep 9th 2008, 07:12 AMNonCommAlg
let be the n-th term of the sequence. it's easy to see that your sequence can be written as: now let:

where and so and: thus :

also we have: thus:

taking square root in (1) and (2) will give us: which means: http://www.mathhelpforum.com/math-he...ags/Canada.gif - Sep 9th 2008, 08:57 AMPlato
Here is another way to generate that sequence: .

That is of course using the ceiling function.

Notice the sequence changes on the triangular numbers, 1,3,6,10,…; . - Sep 9th 2008, 12:31 PMbkarpuz
First of all, I have to say that this is a very good exercise.

Consider the terms marked with , for these numbers, we have

Now, let , where is the smallest number which is not less than and satisfies (in short, lets say property) and is the greatest number which is less than and satisfies property.

Therefore, solving this parabola with, we have . Note that the desired term is the positive one, hence . Thus, the solution is completed when has property.

If does not have property, then we see that , you can easily show that , because of the definitions of , being and the sequezing

Plato's answer is right! (Nod)