You need to use mathematical induction to prove the formula for every positive integer n, hence, you need to perform the two steps of the method, such that:
Step 1: Basis: Show that the statement P(n) hold for n = 3, such that:
2*3^2 > (3+1)^2 => 18>16
Step 2: Inductive step: Show that if P(k) holds, then also P(k + 1) holds:
P(k):2k^2 >(k+1)^2
holds
P(k+1):2(k+1)^2 >(k+2)^2
You need to use induction hypothesis that P(k) holds, hence, you need to re-write the left side, such that:
2*2k^2 > (k+2)^2
4k^2 > k^2 + 4k + 4
k^2 + k^2 + k^2 + k^2 > k^2 + 4k + 4
k^2 + k^2 + k^2 > 4k + 4
k^2 > 4k and 2k^2 > 4 => k^2 > 2
Notice that P(k+1) holds.
Hence, since both the basis and the inductive step have been verified, by mathematical induction, the statement P(n):2n^2 >(n+1)^2 holds for all positive integers n.
Monday, April 16, 2012
Precalculus, Chapter 9, 9.4, Section 9.4, Problem 30
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