Answer:
You proved it's valid for n=5. Presently assume it's valid for some integer n?5. The point is to demonstrate it's valid for n+1. Be that as it may
2n+1=2×2^n>2×n^2>(n+1)^2.
The primary inequality takes after from the acceptance theory and with respect to the second, we realize that (n?1)^2 ? 42 > 2, since n?5. We can grow this inequality (n?1)^2>2 as takes after:
n^2?2n+1>n^2?2n?1>2n^2?2n?1>2n^2>2n^2+2n+1=(n+1^)2,
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