Recent content by John112

I guess they mixed it up then, I guess 31 is the answer from floor 2 to floor 1 then. Thanks for the reply.

Should the right answer to this question(below) be 14 and not 31? because A_{ij}^{k} means number of paths from i to j of length K. So A_{12}^{8} = 14 We then represent the graph as indcidence matrices and go from there on: A = { {0,1,0,0}, {1,0,1,0}, {1,1,0,1}, {1,0,0,0} } A[itex]^{8} = {...

I can now see why sin^{2}(x) should be alternating, but how would I then simplify this algerbracially? how would I simplify ##\left(\sum \alpha_n\right)^2## ? Is using the half angle property for sin^{2}(x) my only method?

since the maclaurin series for sin x is alternating in sign (EQ1) so when you square it to get sin^{2}(x) (EQ2) the (-1)^{n} should become (-1)^{2n} (EQ3) which can be simplified down to (EQ4), but when i checked that series at wolframalpha the series was still alternating like: Why is that? So...

Yeah, I saw that. I wasn't sure whether that pattern of dividing by k factorial would hold for bigger sequences. I was trying to find a proof for that. Thanks for the helpful response. By the way is there already a proof for that general formula?

Are there efficient methods of findings the answer to combination problems like these? The letters A, A, B and C are arranged in any order. How many DISTINCT sequences can we form? If two letters weren't the same, then it would be simple. I can find the answer relatively easily since it's...

Let p[n] be the following property of Fibonacci numbers: p[n]: f_{n+1} * f_{n-1} - (f_{n})^{2} = (-1)^{n}. Prove p[n] by induction. This is the proof I wrote. i used regular induction. Is weak induction sufficient to prove it or do I need to prove this by strong induction? proof: BASE STEP...

To prove: (1 + 2 + 4 + . . . + 2^{n}) + 1 = 2^{n+1} , ∀n ≥ 0 . Basis Step n = 0: LEFT HAND SIDE: (2^{0})+ 1 = 1 + 1 = 2 RIGHT HAND SIDE 2^{0} + 1 = 2 Inductive Step: Assume (1 + 2 + 4 + . . . + 2^{k}) + 1 = 2^{k+1} Then (1 + 2 + 4 + . . . + 2^{k} + 2^{k+1}) + 1 = 2^{k+1} + 2^{k+1} +...

I need a bit of help proving the following statement (n + 2)^n ≤ (n + 1)^n+1 where n is a positive integer. The (n+2) and (n+1) bases are making it hard for me solve this. I tried several time, I can't get the inductive step. Can someone lend me a little hand here? The base case is real...

Thanks for that clear explanation tiny-tim!

Im having a bit of hard time understanding how is that two intergers (a and b) divided by a common divisor (m) have the same remainder imply that the difference of (a and b) will aslo be divisible by m? Essentially what I am asking is: a \equiv b (mod m) \Rightarrow m|(a-b) the "|" means...

\forall a,b,c \in Z a|bc \Rightarrow a|b or a|c To prove this do I only need to show that 'a' can be either even or odd?

\forall b\in Z b \equiv 1 (mod 2) \Rightarrow b^{2} \equiv 1 (mod 8) How do I go about proving this? Can the Chinese Remainder Theorem be used to prove this or is there something easier?

So in a way does it mean: 3k + 1 = a number that is 1 bigger than a factor of 3 or a number that is two less than a factor of 3? 3k + 2 = a number that is 2 bigger than a factor of 3 or a number that is one less than a factor of 3? For k \geq 0, so using n = 3k , we create all numbers like...

So in a way does it mean: 3k + 1 = a number that is 1 bigger than a factor of 3 or a number that is two less than a factor of 3? 3k + 2 = a number that is 2 bigger than a factor of 3 or a number that is one less than a factor of 3?