mathematical induction

[native wren]

Does anybody know how to do these? Pls help
T-T

[young coyote]

Induction is just
-Prove it true for a single case (ie n=1)

• Prove if n=k is true, then n=k+1 is true
  For the first part we can simply plug in n=1
  3^0 >= 1^1 -> 1 >= 1 which is true, meaning n = 1 is true
  Now lets assume we have 3^(k-1) = (k^2), k>=2
  3^k = 3•(3^k-1) >= 3•(k^2)
  (k+1)^2 = k^2 + 2k + 1
  If 3(k^2) >= (k+1)^2 then 3^k >= 3(k^2) >= (k+1)^2 or 3^k >= (k+1)^2, proving the statement for k+1
  3•k^2 >= k^2 + 2k + 1
  2•k^2 >= 2k + 1
  2k^2 - 2k - 1 >= 0
  Using quadratic, we get k>1/2 + sqrt(3)/2
  If n=k is true and k> 1/2+sqrt(3)/2, then n=k+1 is true
  @native wren see what to do from here?

[native wren]

so for the last step of Q1, how would you prove 3(3^(k-1)-k^2)+2k^2-2k-1≥0?