Mathematical Problem! Can anyone prove that this is a infinitely true word problem that extends to 100 steps?

Question

A person can either take one step or two steps at a time to go up a stairs. There will be 1 way to go up one step. There will be 2 ways going up 2 steps. there will be 3 ways going up 3 steps (2-1-1 and 1-2-1 and 1-1-2). There will be 5 ways going up 4 steps (1-1-2 and 2-2 and 2-1-1 and 1-2-1 and 1-1-2). There will be 8, 13,21, 34 , 55 and 89 ways to go up 5,6,7,8,9 and 10 steps in a stair. If you could write those out, there is a pattern. but is there a way to prove that this pattern goes on infinitely? for example n-number of steps?

Bramblyspam has suggested using the induction method. (er... but i dont know how to do that, am not a mathematics major). but well done still. the logic sounds doable.

thanks dennismeng90 for pointing out that mistake. yes. its a typo on my part.

Answer 1

You prove this via induction.
- First, prove that it holds true for the first step. (That's easy).
- Next, prove that if it holds true for the nth step, it must automatically hold true for the (n+1)th step.

If you can prove those two things, then you've proven that the pattern goes on forever.

Answer 2

If you can prove convergence of your infinite series, you can conclude the series is valid toward infinity.

Convergence tests:

Comparison test 1: If ∑bn is an absolutely convergent series such that |an | ≤ C |bn | for some number C and for sufficiently large n , then ∑an converges absolutely as well. If ∑|bn | diverges, and |an | ≥ |bn | for all sufficiently large n , then ∑an also fails to converge absolutely (though it could still be conditionally convergent, e.g. if the an alternate in sign).

Comparison test 2: If ∑bn is an absolutely convergent series such that |an+1 /an | ≤ C |bn+1 /bn | for some number C and for sufficiently large n , then ∑an converges absolutely as well. If ∑|bn | diverges, and |an+1 /an | ≥ |bn+1 /bn | for all sufficiently large n , then ∑an also fails to converge absolutely (though it could still be conditionally convergent, e.g. if the an alternate in sign).

Ratio test: If |an+1/an| approaches a number less than one as n approaches infinity, then ∑ an converges absolutely. When the ratio is 1, convergence can sometimes be determined as well.

Root test: If there exists a constant C < 1 such that |an|1/n ≤ C for all sufficiently large n, then ∑ an converges absolutely.

Integral test: if f(x) is a positive monotone decreasing function defined on the interval [1, ∞) with f(n) = an for all n, then ∑ an converges if and only if the integral ∫1∞ f(x) dx is finite.

Alternating series test: A series of the form ∑ (−1)n an (with an ≥ 0) is called alternating. Such a series converges if the sequence an is monotone decreasing and converges to 0. The converse is in general not true. [See source.]

Answer 3

Just pointing out that the three ways for 3 steps is 1-1-1, 2-1, and 1-2, not 2-1-1, 1-1-2, and 1-2-1.