Alright, because the professor I have right now thinks those are two completely different things. He reads it as "there exists an x such that there exists a y (for that specific x)." This is the second time I had the class (it didn't transfer colleges) and my previous professor said they're the...
d_obj = do = 25 ft = 300 inches
h_img = hi = -3 inches
h_obj = ho = 6 ft = 72 inches
hi/ho = -di/do
-3/72 = -di/300
di=12.5 inches
1/do + 1/di = 2/r
1/300 + 1/12.5 = 2/r
r=24 inches
So the radius of the mirror is precisely 24 inches? That still seems like a rather large convenience store mirror.
The problem:
"A convex spherical mirror is 25 ft from the door of a convenience store. The clerk needs to see a 6 ft. person entering the store at least 3 inches tall in the mirror to identify them. What is the radius of the mirror?"
d_obj = do = 25 ft = 300 inches
h_img = hi = 3 inches...
Yep. I just didn't realize how to change the expression until I looked at examples enough and finally noticed you simply substitute >.< Thanks for your help.
1+2(2+3+4+···+k+k+1)+(k+1+1)
1+2(2+3+4+···+k)+2(k+1)+(k+1)+1
1+2(2+3+4+···+k)+(k+1)+2(k+1)+1
Recall: 1+2(2+3+4+···+k)+(k+1)=(k+1)2-1
(k+1)2-1+2k+2+1
k2+2k+1-1+2k+2+1
k2+4k+3
k2+4k+4-1
(k+2)(k+2)-1
(k+2)2-1
LHS=RHS
Therefore, 1+2(2+3+4+···+n)+(n+1)=(n+1)2-1 for all n within N.
Wait, you just substitute the part of the LHS that existed before adding +1 to both sides for the RHS before adding the +1? If so, this is so much less complicated that I thought. I was under the impression you had to do some kind of summation or use a formula to convert it into the values you...
The problem is that I have no idea where to find formulas to convert the expression 2(2+3+4+···+k) into a term. I know that I need to end up with k^2+4k+3 in order to show the RHS, but it seems like I'll be getting too many values. I've been looking through my book and trying to research it...
Would 2(2+3+4+···+k+k+1) = k(k+2)+k+1?
So it would all equal 1+k(k+2)+k+1+k+1+1
=k^2+4k+4
However, I need it to equal k^2+4k+3
Which I can get if I don't add an extra 1. So I use:
1+2(2+3+4+···+k+k+1)+(k+1)=(k+1+1)2-1
instead of
1+2(2+3+4+···+k+k+1)+(k+1+1)=(k+1+1)2-1
First, by using n=2 for the basis, I confirmed that the equation is true.
Then,
Assume true for n=k
1+2(2+3+4+···+k)+(k+1)=(k+1)2-1 for all k within N
Show true for n=k+1
1+2(2+3+4+···+k+k+1)+(k+1)+(k+1+1)=(k+1+1)2-1
I'm trying to contradict this somehow, and went through with it unsuccessfully...
That's what I was thinking. So since 1 is a natural number, and the problem states that the equation is true for all n in N-the set of natural numbers, should it be seen as 1 not being included in n, and therefore not being in N?