Use induction to prove that: $$2^n n^2 -7$$ for all positive integers n.

Question

Use induction to prove that: $$2^n >n^2 -7$$  for all positive integers n.

anonymous · Answer

so far, I have tested the least positive integer, 1, and it worked. For step two, you assume this is true for k. And step 3, test k+1 right?

hartnn · Answer

correct.

anonymous · Answer

So I've got:  $$2^{k+1} > (k+1)^2 - 7$$

and then I separated things out a bit

$$2^k * 2^1 > k^2+2k+1-7$$

anonymous · Answer

which can also be written: $$2^k * 2 > [(k+1)(k+1)]-7$$
but Im unsure what to do next

anonymous · Answer

oooooooh wait, is it because the k^2 and -7 are from the original equation with justk in it? So I .....do something with them....

hartnn · Answer

yeah , u use the fact that 2^k>k^2-7

anonymous · Answer

Should I maybe, make a chain of inequalities of the right side, to show its smaller than 2^(k+1)

anonymous · Answer

or,divide the right side by two, because if 2^k is > k^2-7, its definitely larger than (k^2-7) / 2 and k and 1/2 would be constants that stay smaller than the exponential on the left? If that makes sense..

anonymous · Answer

$$2^k > (k^2-7)/2 + k + 1/2$$

anonymous · Answer

what if I subrtact them from the left now, and so: $$2^k > 2^k -k + 1/2 > (k^2-7) / 2$$

anonymous · Answer

Has this train completely derailed, or am I getting there?

anonymous · Answer

I feel like the last step is just to say for all positive integers n, 2^(n+1) > 2^n , thus greater than all the stuff on the right.... Aww, crap, could I have just done that from the beginning without all this shuffling around ?

hartnn · Answer

sorry for late response 

i don't used to do it that way
i took 2^k > k^2-7 
multipled 2
2^(k+1) > 2k^2-14
 now we need to show 
2^(k+1) > k^2+2k+1-7 

in short we need to show (*imp*, ask if doubt)
k^2+2k-6> 2k^2-14
can you ?

anonymous · Answer

I cant figure it out, it seems to me like the 2k^2 will make the right side always bigger... Im not sure how to do it.

anonymous · Answer

It just seems like 2k^2 will dominate k^2+2k eventually

hartnn · Answer

oh, wait

a> b 

to prove 
a>c
we need b>c 
is that what i did ?

anonymous · Answer

oooooh, ok, that makes more sense. We show that if $$2k^2-14 > k^2 +2k -6 $$

then $$2^{(k+1)} > 2k^2-14$$ is implied and we are done!

anonymous · Answer

Thanks a lot for the help, I completely understand now :D

hartnn · Answer

oh, i guess i go confused, but u got the point...good
welcome ^_^