Finding the nth term of 3,5,9,15,...

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[Zr_Gaming]

We call this method as 'Method of Difference'. Solved a lot of these while preparing for JEE

[paulortega]

The difference method is very easy when laid out visually in a triangular format.

3 5 9 15 23 33...
2 4 6 8 10
2 2 2 2

You stop at the row that repeats itself. Use the difference values on far left.

f(n) = 3 + 2*(n - 1)/1! + 2*(n - 1)*(n - 2)/2!

f(n) = 3 - n + n^2

[quantumobject]

Wow that's a cool solution .
I just noticed that a1=3
a2=3+2(1), a3=3+2(1+2) ...
So
Which gives us an=3+n(n-1)
But your solution is way better

[Ganondurk]

I used a bit of calculus to solve this problem. If you consider the sequence of differences to be the derivative of the target function, then it's simple to find that it's modeled by the linear function 2n - 1. You then just take the integral and substitute the initial coefficient n_1 for C. The integral of 2n - 1 = x^2 - x + C, and substituting 3 for C gives us a_n = n^2 - n + 3.

You can generalize this method to find nearly any formula for an arbitrary length sequence; just continue finding the difference sequence until the resulting sequence is trivial to model, and then take repeated integrals up from there, substituting in the appropriate coefficient for C at each level.

[이효신-tq]

Without more specific conditions given from the question, I believe these problems do show lack of clarity. We can easily see that there are infinite number of different solutions for this sequence.
For example in polynomials, Let A_n be and k be any arbitrary real(or complex, rational or natural whatsoever). Then, first four components of this sequence clearly satisfies {3, 5, 9, 15, ...} while the sequence itself varys infinitely for any k possible. One can also build a sequence that repeats {3, 5, 9, 15} by combining sines and cosines of (n*pi/2), ((n+1)*pi/2), ((n+2)*pi/2), ((n+3)*pi/2). A word 'sequence' is far too broad.

[matthewfunk]

You can also reason that, since the 1st differences follow a linear pattern, the 2nd differences will be constant, which is analogous to taking two derivatives of a function and getting a constant, which is only possible if the original function is quadratic. Hence, aN is quadratic, so you can create a system of three equations in three unknowns, using the 1st three terms of the sequence, to solve for the unknown coefficients of aN = C1N^2:+ C2N + C3.

[Vladimir_Pavlov]

You can draw a parabola on a plane, and in the only way, through any three points that do not lie on the same straight line.
Let's draw a parabola a (x)= Ax^2 +Bx +C through points (1;3), (2;5) and (3;9).
We obtain a system of equations :
A+B+C=3, 4A+2B+C=5, 9A+3B+C=9, solving which, we easily find A=1, B=-1, C=3,
that is, a(n)=n^2-n+3. In this case, a (4)=15.
But what if the fourth term of the series is, for example, 16 ?
Then a (n)=An^3+Bn^2+Cn+D. Let's write a system of 4 equations, from which we find
a (n)= (1/6)n^3+(5/6)n^2+2.)))

[jmart]

Once you notice that the difference between two terms is 2n then you can be sure that the general term will be of the form an^2+bn+c
So you get three equations

a+b+c = 3
4a+2b+c=5
9a+3b+c=9
Which are easily solved giving a=1, b=-1 c=3

[kaslircribs]

Very interesting your method of solving the problem. I use the differences and determine the missing coefficients. Thank you very much sir.

[nalapurraghavendrarao]

Use method of succesive differences. If second order differences are constant, then 2 nd degree equations fits.take it as an^2+bn+c and substitute n=1, 2, 3...You get 3 equations to solve

[vuyyurisatyasrinivasarao]

Ur method super great mathematician ur sir some special in ur teaching....good that new....

[alfreddino]

Disagree.. According to the latest video of Micheal Penn (Math from Iceland), it is a sequence of 3^n * 5^m. haha~~ First few terms are: 3, 5, 9, 15, 25, 27, 45, 75, 81, 125, ... haha~~ What a coincidence!!!

[jarikosonen]

It seems to work this way and also if there is more initial terms it can be more certain how it goes. What If this was function of m & n instead of n only?

[adandap]

Nice. You could also use this problem to introduce the idea of a 'summing factor' in analogy to the 'integrating factor' in first order DEs.

[borisedgartribenouyuni]

I think it could be any number and Sybermath will explain us why. (Probably polynomial functions)

[JalebJay]

I looked at the problem and thought of the alternating sequence of:
a_1 = 3
a_n = 9/5 a_{n-1} if n>1 is odd
a_n = 5/3 a_{n-1} if n>1 is even.

Which turns into:
a_{2n-1} = 3^n
a_{2n} = 3^(n-1)*5

[advaykumar]

You can also do it by first seeing difference are in ap then assuming nthe term by an^2+bn+c then putting n= 1 2 and 3 and setting it to 3, 5 and 9 respectively

[yanmich]

I saw another recurrence relation: a_{n+2} = a_{n+1} + a_{n} + 1

[ethang]

Don't know about anyone else but this is GCSE stuff in England.

[mputuchimezie]

Great video.
Sir Please how do you find an nth term formula for a sequence like 14, 17, 23, 35, 59 ?