How are Equations for Different Series Generated?

[AdityaDev]

Given series:1,2,5,12,25...

How did they get :$T_n=a(n-1)(n-2)(n-3)+b(n-1)(n-2)+c(n-1)+d$

And for series like 3,7,13,21,...
they have given $T_n=an^2+bn+c$

How do you get these equations?

 

[Mark44]

Given series:1,2,5,12,25...
View attachment 80673
How did they get :$T_n=a(n-1)(n-2)(n-3)+b(n-1)(n-2)+c(n-1)+d$

And for series like 3,7,13,21,...

Based on the image, this should be 1, 3, 7, 13, 21, ...

they have given $T_n=an^2+bn+c$

How do you get these equations?

If the first consecutive differences happened to all be the same constant, the solution would have been a linear (first-degree) polynomial. If the second differences turned out to be a constant (as in the sequence 1, 3, 7, 13, 21, ...), the solution would be a quadratic (second-degree) polynomial, which you could write as T_n = a(n - 1)(n - 2) + b(n - 1) + c.

In your problem, which involves the sequence 1, 2, 5, 12, 25, 46, ... the third differences are all 2, so the solution will be a cubic (third-degree) polynomial, which they write as $T_n = a(n - 1)(n - 2)(n - 3) + b(n - 1)(n - 2) + c(n - 1) + d$
What they do after this is to solve for the coefficients a, b, c, and d, noting that T₁ = 1, T₂ = 2, T₃ = 5, and T₄ = 12. IOW, T_n is just the n^th term in the sequence.

As an aside, 1, 3, 7, 13, 21, ... is a sequence of numbers, not a series. In a series, all the numbers are added together to produce a sum.

I'm pretty rusty on this stuff, as it has been many years since I did anything with difference equations. Nevertheless, I was able to get the coefficients a, b, c, and d, and was able to get the correct value for T₅, so I think I'm on the right track.

 

[AdityaDev]

Why is it T_n=a(n-1)(n-2).. why not $T_n=a(n-4)(n-8)+b(n-25)+c$

 

[Mentallic]

If you ignore the way they've expressed it for the moment, you can still agree that it's a cubic, correct? It could just as easily be expanded and turned into the general cubic form
$$T_n=An^3+Bn^2+Cn+D$$
where A,B,C,D are likely going to be different constants to a,b,c,d but equate to the same cubic.

Now, why was that particular form chosen?

Well, picking n=1 gives us $T_1=a(1-1)(1-2)(1-3)+b(1-1)(1-2)+c(1-1)+d$. Now, notice that any (n-1) factor gives us 0, hence we end up with $T_1=a*0+b*0+c*0+d=d$ so with this form, we can easily find d, as opposed to having the general cubic with coefficients A,B,C,D that I had shown above. In that case, $T_1=A+B+C+D$ and we're hardly any closer to finding the solution.

So, we have $T_1=d=1$ and we've already knocked one of the coefficients out of the way. $T_2$ turns out to give us a constant as well by following a similar idea.
$$T_2=a(2-1)(2-2)(2-3)+b(2-1)(2-2)+c(2-1)+d$$
and again, notice that (n-2) factors would equal zero, hence we end up with $T_2=a*0+b*0+c+d=c+d$. But we already found d=1, so $T_2=c+1=2$ hence $c=1$. For n=3, you'll end up with just b,c,d and can solve for b easily since you know c,d.

 

[AdityaDev]

Thank you @Mentallic

 

[Mentallic]

Thank you @Mentallic

You're welcome, and good luck with your studies!