Write a recursive formula for the sequence 15, 26, 48, 92, and 180 …then find the next term.
\[(x - 2) \times 2\] sry not sure about the recursion part unless you want c++ :)
This is probably really easy but I need help understanding the concept of a recursive formula
a1 = 15, an+1 = an + 2n-1×11 To get the next term an+1 from the previous term an we must add 2n-1×11. the recursion formula is a1 = 15, an+1 = an + 2n-1×11 So to get the 6th term, a6, from the 5th term,a5, I substitute 5 for n in an+1 = an + 2n-1×11 a5+1 = a5 + 25-1×11 a6 = a5 + 24×11 Now we substitute 180 for a5 a6 = 180 + 24×11 a6 = 180 + 16×11 a6 = 180 + 176 a6 = 356
I understand it all now but the n-1...it never makes sense, what do I do with the 1?
an-1-4, where a1=15; 356 ...I am suppose to have this kind of answer but when I plug the numbers in, it isn't right.
a_1 = 15 ; a_n = a_(n-1) + 5.5 • 2 ^ (n-1) ░░░░░░░░░░░░░░░░░░░░░░░░░░░░ a_6 = a_5 + 5.5 • 2 ^ 5 ..... = 180 + 5.5 • 32 ..... = 180 + 176 ..... = 356 Is all i did to get it.
Where did the 4 come in my answer then? ....you see I am suppose to correct this answer but I don't have any clue at all....
I don't see where you got it at all.
It was the correct answer to this problem and don't have the slightest clue what I to do with the n-1...it's in all of my problems and without it, I get correct answers but if I use n-1, I get the wrong answers.
oops, I made a mistake the answer should look like\[2a _{n-1}-4\]
Yes.
Wow its like 11:14 here :D. u?
Same lol
I live in Fl.
Do you know what I do with the n-1? lol
Not sure. i am bit sleepy but i don;t want to go to bed..
@Mertsj
\[a _{n-1}\] is the term just before the nth term. Typically the sequence would look like:
\[a _{1}, a _{2}, a _{3}, ...a _{n-2}, a _{n-1}, a _{n}\]
Ohhh, thank you so very much!
So we could say \[a _{n}= 2a _{n-1}-4\]
yw
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