# Method of difference

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1. Mar 19, 2015

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?

2. Mar 20, 2015

### Staff: Mentor

Based on the image, this should be 1, 3, 7, 13, 21, ...
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 Tn = 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, Tn is just the nth 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.

3. Mar 20, 2015

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

4. Mar 20, 2015

### 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.

5. Mar 20, 2015