how can we get next number in this series 4, 21, 143, 1061, 8363 ......
i can make you a formula that will work if you want <.< It may not be what the creator of the problem intended though.
kk, pls do help me
alright, one sec.
and ya also teach me to make those kind of formulas so that next time i can do any such kind of problem
you can change the value of n at the end of the function to see that it works, make n = 6 to see the 6th term. The method for creating these involves a bit of linear algebra, if you arent very comfortable with Linear Algebra it might be a little complicated.
yeah show me too, although that has to be way uglier than intendend
good joe math , keep it up
i am good at algebra , teach me how to made that function
joe please also tell me about some software that i can use to plot graphs and functions and to get functions from plotted graphs ?
Alright. The idea is two-fold. 1st, we are going to abuse the fact that the tail of the sequence is unknown. We dont know if this sequence converges, diverges, whatever. 2nd, since we dont know about the part of the sequence beyond the 5th term, wouldnt it be nice if the 6th term was some linear combination of the first 5 terms? Who is to say it is or isnt?
ok , what next?
So, (without going into too much detail, since this might be my undergraduate thesis <.<), Step 1, pick 5 integers that are all different. I picked -2, -1, 1, 2, 3. Step 2 calculate their powers and make the, into a square matrix. You end up with: -2 -1 1 2 3 4 1 1 4 9 -8 -1 1 8 27 16 1 1 16 81 -32 -1 1 32 243 Lets call this matrix A
i don't mean to interrupt so i will let you continue, but i can tell you what this sequence represents if you like
Step 3: Solve this matrix equation for x: \[Ax = b\]where b = (4, 21, 143, 1061, 8363)
the vector x will be the coefficients of the powers of the 5 numbers you chose.
y we did that ?
so after i chose the numbers -2, -1, 1, 2, 3, i knew the solution was going to be in the form: \[c_1(-2)^n+c_2(-1)^n+c_3(1)^n+c_4(2)^n+c_5(3)^n\] Solving that matrix equation tells me what the c's are.
hey, this kind of thing was not taught to us ?? it is given in some book or you made it ?
can you teach me applying that on some easy series ??? please
4, 21, 143, 1061, 8363, 68906, 586081, 5096876,
It wasnt something taught in a class per se...its kinda like working linear recurrences backwards...i did get the idea on my own. I doubt im the first person to think about it though, its not a super complicated idea.
saifoo it is your bed time!
@joe you know what these numbers are?
not at all! is there some pattern? now that i have this way of creating sequence formula's, ive given up on recognizing patterns lol
hey , u r bad? u scared me
@joe hey, please teach me by applying this on some easy series , like some arithmetic series please
if you had one you are famous. and how divanshu is supposed to come up with the next one is beyond me. an is the number of primes of n digits. 1 digit, 2 primes, 2 digits, 21 primes, 3 digits, 143 primes which makes me suspect that the question is a joke
should be 1 digit, 4 primes
oh woah <.< thats ridiculous. @divanshu if the series is arithmetic or geometric, then there are already formulas for the nth term. I use this method when i cant figure out what the formula is.
hey , the question was in my college exam , i can show u paper ??
in fact i see that the question has been asked and answered an hour ago
@joe that whats i am saying , to apply it on some simple thing about which i can cross check my answers?
ya, i asked it an hour ago and no one answered me completely
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