Hello,
I need some more clarification on a question I asked earlier.
Here's the problem:
Prove by induction on n that, n^3<=2^n for n>=10 a positive integer.
First, Thank you Pascal and rhsaunder...
Pascal. I am being taught to solve these proofs very similar to the way you solved the problem. I need some clarification if I can.
Here is what I have so far.
P(n): n^3 <= 2^n for n>=10 a positive integer.
Step 1: Let n=10 so I get 10^3 <= 2^10 which is 1000<=1024
Step 2: Assume P(k) is true for n=k
So, k^3<=2^k (Ind. Hyp.)
Step 3: Must show P(k+1) is true. Let n=k+1
So, (k+1)^3<=2^(k+1)
We know k^3<=2^k
From here 2k^3<=2^k*2
Which is 2k^3<=2^(k+1)
Then:
(k+1)^3 = K^3+3k^2+3k+1=K^3+(9K^2+9k+3)/3
From here I get lost when I introduce n>=10.
Need clarification on this - Since n≥10, 9n² < n*n²=n³, and likewise 9n < n³ and 3 < n < n³.
Then:
n³ + (9n²+9n+3)/3 < n³ + (n³+n³+n³)/3
n³ + (n³+n³+n³)/3 = n³+n³ = 2n³
Thank you in advance!
2007-02-09
06:09:27
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2 answers
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asked by
MathStudent3
1
in
Mathematics