Recent content by nastygoalie89

I just don't see how to get from k to (k+1)k^k

how do I get to k(k+1)k ≤ (k+1)k+1 from k<k+1, without multiplying (k+1)^k?

i think i have it. k! < k^k, multiply by (k+1) k!(k+1)= (k+1)! < (k+1)k^k and since k<= k+1, multiply by (k+1)^k this gives us (k+1)k^k < (k+1)^(k+1) we proved (k+1)! < (k+1)k^k < (k+1)^(k+1) therefore, (k+1)! < (k+1)^(k+1)

ah i have no idea how to prove that without starting with 'q'

1. k!< k^k 2. multiply by (k+1) to get k!(k+1) <= (k+1)k^k 3. (k+1)k^k <= (k+1)^(k+1) multiply by k to get (k+1)k^(k+1)<=(k+1)k^(k+1) therefore, k!(k+1) <= (k+1)k^(k+1), divide by (k+1) k! <= (k+1)^k

1. k!< k^k 2. multiply by (k+1) to get k!(k+1) < (k+1)k^k 3. k!(k+1) < (k+1)k^k < (k+1)^(k+1) 4. therefore, k!(k+1) < (k+1)^(k+1) I feel I am begging the question at step 3 here

how do you get (k+1) k^k? isn't it (k+1)(k+1)^k?

so it would be: k!<=k^k multiply by (k+1) on both sides (k+1)! = k!(k+1)<= (k+1)^k (k+1) since k is an integer k+1 is an integer am I closer?

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so it would be k!(k+1) <= (k+1)^k + (k+1) ?

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