How to find the nth term of this sequence -11, -13, 7, 61, 161 I've solved one part of the question, which is that the numbers have a relation of 12. But after that im lost.
-12*
what do you mean by they have a relation of -12?
sorry i actually meant 12, made a calculation mistake. What i mean is that when simplified, the pattern gives a result of 12
Im afraid I don't follow that.. Simplified?
hopefully this will explain it better, sorry for bad drawing.
Hopefully that was readable
yes
so what would be the answer
I cant figur out this one Lets see if its on the encyclopoedia of series
https://oeis.org/search?q=-11%2C-13%2C7%2C61%2C161&language=english&go=Search Nope
So is it not possible to solve?
maybe
But because its not on the list doesn't mean its impossible.
i have an idea, but i dont know if it is correct, but we were learning quadratic sequences in class so this could be quadratic sequence and the quadratic formula for this sequence could be an^3 + bn^2 + cn + d Does this make sense? im new to this topic
@welshfella
Ok I have a solution
we will do it by newton's forward interpolation formula..
uhmm okay, but please keep in mind that i am in 9th standard so hopefully the formula isnt too complicated
the general formula, will be 2(n-1)(x-2)(x-3)+11(n-1)(n-2)-2x-9
oh.. replace those x by n
f(n)=2(n-1)(n-2)(n-3)+11(n-1)(n-2)-2n-9
check if the values are correct or not... put n=1, n=2, etc
ok i will try it, thanks for the help!
will you not learn how I got it?
I've never heard of tht newton formula. Must look it up.
But check , I got the formula right, just put the values @welshfella
it works for n = 1,2 and 3 - giving -11,-13 and 7
check kor n=4 and 5 also
im going too now
works for n = 4
Yes also for n = 5
good job
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