Prove by induction n^2 > 5n+10 for n > 6. 1. Base case: 2. State the inductive h

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1. base case: proving for n = 7 l.h.s = 7^2 = 49 r.h.s = 5*7 + 10 = 45 hence it is true for base case. 2. inductive hypothesis is if (the statement is true for a k > 6 implies that it is true for k+1) then the statement is always true. 3. we have to show that given k^2 > 5k + 10 ==> (k+1)^2 > 5*(k+1) + 10 = 5k + 15 4. proof: given k^2 > 5k + 10 now take (k+1)^2 (k+1)^2 = k^2 + 2k +1 as k^2 > 5k + 10 k^2 + 2k +1 > 5k + 10 + 2k +1 (k+1)^2 > 5k + 13 + 2k here k>6 so 2k>12 ==> (k+1)^2 > 5k + 13 + 2k ==> (k+1)^2 > 5k + 13 + 12 ==> (k+1)^2 > 5k + 25 it is quite obvious that (k+1)^2 > 5k + 15 i.e (k+1)^2 > 5*(k+1) + 10 hence as the inductive hypothesis is proved, the statement is always true.

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