Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
amistre64
Group Title
Think of 3 rules, such that they generate 2,3,6 as the first 3 terms of a sequence.
 one year ago
 one year ago
amistre64 Group Title
Think of 3 rules, such that they generate 2,3,6 as the first 3 terms of a sequence.
 one year ago
 one year ago

This Question is Closed

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
1/2 3/2 4 i came up with this one:\[\frac{1}{2}\left(\frac{n1.5}{n1.5}+\frac{3n7.5}{n2.5}\frac{8n280}{n3.5}\right)\]
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
lol, i forgot to delete my reference scribbles
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
i was also considering coming up with something such that:\[\int_{0}^{n}f(x)dx=2,3,6~for~n=1,2,3\]
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
the first one i thought of was\[a_n=n^22n+3\]
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
dw:1346159138909:dw hmm; such that the first area is 2, and the second is 1, and the third is 3
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
if i could determine the slopes in that figure, i could integrate the function to get what im thinking of :)
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
dw:1346159507196:dwba=1\[\int_{a}^{b}k~dx=F(b)F(a)=1\]F(b)=4, F(a)=3
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
err, F(a) = 3
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
dw:1346159801695:dw 0 to b , b<1 ... its like right on the tip of my brain
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
dw:1346160036988:dw at a, A=2, at (ba)/2, A=0 so there has to be a sweet spot between a and the midpoint that satisfies my quandry
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
if b= 3/4; and a = 1/4 \[\frac{4}{3/4}x=\frac{16}{3}x; \int\to\ \frac83 \left((\frac34)^2(\frac14)^2\right)\]\[\frac83 \left(\frac{91}{16}\right)=1\frac13\] so its between 1/4 and 1/2 :)
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
dw:1346161272353:dw hmmm
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
not B=8, 4+4=8/2=4; that triangle area there is B=4
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
if i guess that: (4+d)/2 = 3; d=2 mxd 4+2 = 6, 6 in 1 = 6x, 6x2 3x^22x, (0,1) = 32=1
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
dw:1346162349632:dw
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
2,3,6 is my goal; and i wonder if alternating signs, and doubling would give me my needed slopes, making the next one 12 12x2; 6x^22x = 4, not 3 it was a nice guess
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
int mx2 , [0,1] m/2  2 = 3 m/2 = 5 m=10
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
dw:1346163082261:dw
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
abc = 4 abc=6 a+bc=10 rref{{1,1,1,4},{1,1,1,6},{1,1,1,10}} 5,8,7 f(x) = 5x1+8x27x3 and since n=0 = 10; lets add 10 to it to move it to the proper spot \[\int_{0}^{n}f(x) = 5x1+8x27x3+10~dx=\{2,3,6,...\}~:~n=1,2,3,...\]
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
\[\int_{0}^{n}f(x)~dx\]\[\int_{0}^{n} (5x1+8x27x3+10)~dx=\{2,3,6,...\}~:~n=1,2,3,...\] that fixes the typos
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
ima goofy goober!!! lol
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
i know right, i should be soo ashamed
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
i guess my idea of a second degree polynomial is too prosaic. you probably did that first anyway
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
maybe change it to "find the most arcane sequences whose first three terms are ..."
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
i did .... and my second original idea was:\[a_{n+1}=a{n}*a_{n1}~:~a_1=2, a_2=3\]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
that is a good one
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
if by arcane you mean deviously clever ... i agree
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
i really meant "off the wall" not arcane
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
i wonder what the math teachers gonna give me as a grade on the homework this year; or if shell even read it
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
you will get a gold star for cleverness
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
i gotta hobble over to physics and 2 other classes for awhile; yall try not to break stuff while im gone
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
\[\large{s_n=2n\frac n4\left(1+(1)^n\right)}\] \[s_1=2\frac 14\left(1+(1)^1\right)=20=2\] \[s_2=2\times2\frac 24\left(1+(1)^2\right)=4\frac12(2)=3\] \[s_3=2\times3\frac 34\left(1+(1)^3\right)=60=6\]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
\[a_n=(1)^{n}n^2\cos(\pi n)+(1)^{n+1}2\cos(\pi n)+(1)^n3\cos(\pi n)\]
 one year ago

eliassaab Group TitleBest ResponseYou've already chosen the best response.0
Why is interesting? I do not know. Here is my simple contribution. \[ a_1=2\\ a_2=3\\ a_n=a_{n1}a_{n2},\quad n>2 \] \[ a_1=2\\ a_2=3\\ a_n=a_{n1}+a_{n2}+1,\quad n>2 \] \[ a_1=2\\ a_2=3\\ a_n=a_{n1}a_{n2}+5,\quad n>2 \]
 one year ago

eliassaab Group TitleBest ResponseYou've already chosen the best response.0
Here are two more. You can have infinite such rules. \[ a_1=2\\ a_2=3\\ a_n=2 a_{n1} ,\quad n>2 \] \[ a_1=2\\ a_2=3\\ a_n=3 a_{n2} ,\quad n>2 \]
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
the question was so bland that i figured id see if there were more interesting ways to develop a sequence that began with the given digits.
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
if my teacher insists that i have to do the homework for a grade .... im at least going to try to get some fun out of it :)
 one year ago

Herp_Derp Group TitleBest ResponseYou've already chosen the best response.1
\[a_n=\Pi (n)\Gamma (n) +2\] \(\Pi (n)=n!\) and \(\Gamma (n) =(n1)!\) but it just looks so much cooler that way.
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
lol, awesome
 one year ago

Herp_Derp Group TitleBest ResponseYou've already chosen the best response.1
This one is just... idk \[\large a_n=\left\lfloor\sqrt[n]{n+\Big\lfloor e^{\frac{1}{2}n^2}\Big\rfloor}\right\rfloor +\left\lfloor\int_0^n \frac{\frac{1}{2}xe^{\sqrt{x}}}{\sin (x)}\,\text{d}x\right\rfloor\]
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.