a.)
n=1: 1 n=6: 21
n=2: 3 n=7: 28
n=3: 6 n=8: 36
n=4: 10 n=9: 45
n=5: 15 n=10: 55
n=2: 3 n=7: 28
n=3: 6 n=8: 36
n=4: 10 n=9: 45
n=5: 15 n=10: 55
b.)The pattern is that you just add n to the previous sum. I also noticed that every other number n, results in a multiple of your n. Ex. n=9: 45, n=5: 15
c.)I'm almost positive there is a formula for this, because there are formulas for almost every pattern. This pattern also increases by the next number in numerical order.
d.)After thinking for awhile, I came up with this formula:
n=x
(n-1)+x
(n-1)+x
Only thing bad about this formula is you couldn't just plug in 1,000,000 to the formula. You would have to know what n=999,999 is. So, this formula is kind of useless, but its the best I could think of.
At some point can you respond to Prompt #1 as well? I didn't make it very clear that I intended for both to be answered, but as a result that will have you caught up for this week and next week already (and your answer to #1 doesn't have to be very long.) Sorry for the inconvenience.
ReplyDeleteAs for your answer, it's really exactly what I was looking for.
Part a and b are fine (particularly that you pointed out the multiples of n appearing when n is odd which makes me realize you actually thought about this for more than a couple of seconds.
Part c isn't entirely true, I can think of a ton of sequences with no pattern to them, but nonetheless your answer is sufficient.
For part d instead of having "n" and "n-1" you should have a_n (a sub n) and a_(n-1) (meaning the sequence at that n, and not the number n itself.) I see what you're getting at though, but (like you said) this is, albeit correct, a dependent definition (in that you have to know what the value at n-1 is before you can figure out n.) Either way though, your post is fine, I just wanted to add a few comments.