Question

t sub n = n + 1/n is arithmetic, right? The other choices are geometric and neither.

Answer 1

Oh, and I have to write the first four terms...

Answer 2

So to sum up, the sequence is neither arithmetic nor geometric

Answer 3

ok, thanks!

Answer 4

sure thing

Answer 5

Wouldn't term 2 be 5/2?

Answer 6

oh right, made a typo and mixed up d and the second term

you are correct,


T2 = 2+1/2 = 5/2

Answer 7

Ok, could you check term one as well? I'm getting 2.

Answer 8

yeah another very silly typo, 1+1/1 = 1+1 = 2...lol not sure how i missed that...

Answer 9

ah, thanks:)

Answer 10

The first four terms are
T1 = 1+1/1 = 2
T2 = 2+1/2 = 5/2
T3 = 3+1/3 = 10/3
T4 = 4+1/4 = 17/4

So the first four terms are
T1 = 2
T2 = 5/2
T3 = 10/3
T4 = 17/4

Answer 11

I have another problem like this one... t sub n = 16 times 2^2n

Same thing, figure out what kind it is, if any, and give the first four terms.

Answer 12

T1 = 16*2^(2*1) = 16*2^2 = 16*4 = 64
T2 = 16*2^(2*2) = 16*2^4 = 16*16 = 256
T3 = 16*2^(2*3) = 16*2^6 = 16*64 = 1024
T4 = 16*2^(2*4) = 16*2^8 = 16*256 = 4096


So the first four terms are

T1 = 64
T2 = 256
T3 = 1024
T4 = 4096


My initial guess is that this is a geometric sequence (since we have an exponential expression)


If this is case, then the following must be true


r = T2/T1
r = T3/T2
r = T4/T3

where r is some number that is the same for all 3 equations


So


r = T2/T1 = 256/64 = 4, which makes r = 4

r = T3/T2 = 1024/256 = 4, which makes r = 4, so far so good, last one

r = T4/T3 = 4096/1024 = 4, which makes r = 4


The common ratio is r = 4 every time. So this is indeed a geometric sequence.


Alternatively, we can write T[n] = 16*2^(2n) into the form

T[n] = 16 * 2^n * 2^n

T[n] = 16*(2*2)^n

T[n] = 16*(4)^n


and we can say that the last equation shown above is in the form T[n] = a*r^n where r = 4. This further confirms that this is a geometric sequence.

Answer 13

thanks!!

Answer 14

you're welcome