# 1.2 Number Puzzles and Sequences - Excercise 1c

Posted by Gustav Bertram
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 1.2 Number Puzzles and Sequences - Excercise 1c May 01, 2008 05:18PM Registered: 11 years ago Posts: 41 Rating: 0
Hi,

I did the exercises for this section, and I derived a different recursive formula for 1c. I was wondering if I could prove the book answer and my answer are equivalent.

Given the sequence (2, 5, 10, 17, 26, 37...), describe the recursive formula.

I said:
a sub n = (sqrt(a sub n-1 - 1) + 1)^2

The book said:
a sub n = a sub n-1 + (2n -1)

I know both are correct, but is there a way to prove it?

Also, this notation is somewhat clumsy. Is there any way we can attach images to messages, or use TeX or another standard ascii maths notation?
 Re: 1.2 Number Puzzles and Sequences - Excercise 1c May 12, 2008 10:41AM Registered: 11 years ago Posts: 31 Rating: 0
The sequence is 2,5,10,17,26,...

If I understand it correctly, you give the recursive formula as

a sub n = (sqrt(a sub n-1 - 1) + 1)^2

Now take n = 3. (I use indexing to make reading easier.)

a(3) = (sqrt(a(2) - 1) + 1)^2
= (sqrt(5-1) + 1)^2
= (2+1)^2
= 9

which is not correct.

The correct recursive formula is a(1) = 2
and a(n) = a(n-1) + (2n - 1)

E.g. a(3) = a(2) + (2.3 - 1)
= 5 + 5
= 10
 Re: 1.2 Number Puzzles and Sequences - Excercise 1c May 15, 2008 11:54AM Registered: 11 years ago Posts: 41 Rating: 0
Hi,

My bad - I typed in my formula incorrectly. 1 should be added to the result. My correct formula is:

a_n = (\sqrt{a_{n-1} - 1} + 1)^2 + 1

Notation notes:
I'm using AMS-LaTeX notation.
_ is a notation for subscript
a_1 would be a sub 1.
\sqrt[x]{y} is the x'th root of y. With [] omitted, square root is assumed.
\sqrt{2} would be the cube root of 2.
{} are non-displaying grouping indicators.
^ is a notation for superscript.
a^2 would be a to the power 2.

When n = 3:

a_3 = (\sqrt{a_2 - 1} + 1)^2 + 1
= (\sqrt{5-1} + 1)^2 + 1
= (2 + 1)^2 + 1
= 9 + 1
= 10

Which is correct.

Take n = 4:
a_4 = (\sqrt{a_3 - 1} + 1)^2 + 1
= (\sqrt{10-1} + 1)^2 + 1
= (3 + 1)^2 + 1
= 16 + 1
= 17

And so on.

Is there some way I can prove that my formula is equivalent to the given formula?
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