Suitable method for Induction (1 Viewer)

[frog1944]

Hi,

Is the following a suitable method for induction? (For the purpose of brevity, I'm going to skip the testing and the conclusion)


Thanks

 

[InteGrand]

Hi,

Is the following a suitable method for induction? (For the purpose of brevity, I'm going to skip the testing and the conclusion)


Thanks

 

[frog1944]

Ok, though if I just used the graphical method would it still be ok (For the HSC)?

 

[InteGrand]

Ok, though if I just used the graphical method would it still be ok (For the HSC)?

I guess so, but it's probably safer to use the algebraic method.

 

[frog1944]

Ok, thanks InteGrand

 

[aroon]

If the question doesn't explicitly ask for MI (e.g prove than 2^n>/= n^2 (n>/=4)(idk how to use LaTeX and ceebs learning)) can you just draw a graph of the two functions and say that due to the gradient of n^2 always being less than 2^n, 2^n will always be larger?

 

[pikachu975]

Just about the OP's question, is there a way to do that by showing 2^n - n^2 >= 0 (minusing n^2)? Thanks!

 

[Shadowdude]

If the question doesn't explicitly ask for MI (e.g prove than 2^n>/= n^2 (n>/=4)(idk how to use LaTeX and ceebs learning)) can you just draw a graph of the two functions and say that due to the gradient of n^2 always being less than 2^n, 2^n will always be larger?

no, because the gradient of x^2 is larger than the gradient of 500+x, but x^2 isn't always larger

you need to specify an initial point and a range of values

Just about the OP's question, is there a way to do that by showing 2^n - n^2 >= 0 (minusing n^2)? Thanks!

probably, but i envision it'd be more difficult