Induction help (1 Viewer)

[Chicken Burger]

Hi need help with the following question: 2^n > 3n^2, for n is greater than or equal to 8.

Could I also see full working out, so the conclusion and stuff? thanks guys

 

[chris_power_96]

Step 1) Prove for n=8
LHS:2^8=256 RHS: 3(8)^2=192, therefore 2^n>3n^2 is true for n=8

Step 2) Assume true for n=k
ie 2^k>3k^2

Step 3) Prove for n=k+1
from step 2, 2^k>3k^2
therefore, 2(2^k)>2(3k^2)
therefore, 2^(k+1)>6k^2
therefore, 2^(k+1)>3(k+1)^2+3k^2-6k-3>3(k+1)^2 [as (3k^2-6k-3) is positive, as k is above or equal to eight]
therefore, 2^(k+1)>3(k+1)^2

Step 4) Therefore true by mathematical induction

 

[Chicken Burger]

Step 1) Prove for n=8
LHS:2^8=256 RHS: 3(8)^2=192, therefore 2^n>3n^2 is true for n=8

Step 2) Assume true for n=k
ie 2^k>3k^2

Step 3) Prove for n=k+1
from step 2, 2^k>3k^2
therefore, 2(2^k)>2(3k^2)
therefore, 2^(k+1)>6k^2
therefore, 2^(k+1)>3(k+1)^2+3k^2-6k-3>3(k+1)^2 [as (3k^2-6k-3) is positive, as k is above or equal to eight]
therefore, 2^(k+1)>3(k+1)^2

Step 4) Therefore true by mathematical induction

Thanks, but can you explain to me how and why you got the line in bold?

 

[Chicken Burger]

wait nvm i got it thanks for the help chris