Show that n2>2n+1 for n≥3 using mathematical induction method.?

This EXACT problem was asked about two hours ago. I find it hard to believe you coincidentally asked the exact same math problem as someone else in only a two hour time frame.

Its obvious to me this is homework... that you are desperately trying to bust out an hour before school. Your classmates were smart enough to cheat earlier in the day.

What I find amazing, however, is that its so obvious. Of all the worlds population that could theoretically be asking math questions right now, I pick up on this.... Really, I doubt that two classes spread out across the country could have the exact same problem assigned due in the exact same morning. It seems more likely to me that a SINGLE CLASS has gotten the bright idea to do their homework with YA.

Is it a cheating strategy that some geek introduced to all the lazies that attend your one particular class?

The next term of the sequence, i.e the (n+1)th term 1, 3, 5, ..., (2n-1) which is summed is (2n+1), now with n=1 the relationship, 1 + 3 + 5 + ... +(2n-1) = n^2 -----------(1) holds obviously since both sides are 1. Now say (1) holds for n = k for some positive integer k, then, 1 + 3 + 5 + .. .+ (2k-1) = k^2 add the next term (2k+1) to both sides, then; 1 + 3 + 5 + . . .+ (2k-1) + (2k+1) = k^2 +(2k+1) = k^2 + 2k +1 the rightmost side is k^2 + 2k +1, familiar? that's because it is the expansion of (k+1)^2 So we have the relationship true for n=k+1 as well as the originally considered n=k, Now with n=k=1 we know it's true as we first discussed. So it's true for n = 1+1 = 2. And from that it's true for n= 2+1 = 3 and so on for all positive integers and the result is proven by mathematical induction on all positive integers n. QED Hope this helps!!! First answer gives an effective classical method but does not qualify as induction!

Proof: Using Math Induction

P(n): n2 > 2n+1, for n>=3

Base Step: n = 3

3^2 > 2(3) + 1

9 > 7

-> Base Step holds.

Inductive Step: Suppose that the proposition P(n) is true. We must show that P(n+1) is also true.

(n + 1)^2 = n^2 + 2n + 1

> 2n + 1 + 2n + 1 [Using the proposition P(n)]

> 2(2n) + 2 [since 2A > A + 1 holds for A = 2n and n>=3]

> 2(n+1) + 1

-> The Inductive Step holds.

Thus, by the Principle of Math Induction, the inequality holds true for values n>=3.

1^2 = 1 < 3 = 2(1) + 1

2^2 = 4 < 5 = 2(2) + 1

3^2 = 9 > 7 = 2(3) + 1

4^2 = 16 > 9 = 2(4) + 1

5^2 = 25 > 11 = 2(5) + 1

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hence by induction n^2 > 2(n) + 1 when n = 3 or n > 3.