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MATH TRICK-11

It can be rather told as my observation in school days on the chapter Sequence.
Before telling the trick let me explain some terms:



A recurrence relation is a mathematical equation which recursively defines a sequence,where one or more initial terms are given & each further term of the sequence is defined as a function of the preceeding terms.

Ex. The Fibonacci Numbers:
0,1,1,2,3,5,8,13, . . . . .
Here  fn = fn-1 + fn-2  with given values 
fo = 0 , f1 =1 .
Remark: Difference equation is a special type of recurrence relation. There are several techniques to solve recurrence relations. For instance, you can use linear algebric method, z- transformation method,etc.

My Observation-1:-  In most of the problems you can see the recurrence equation is given with one or more initial terms. Then if you use basic algebra & try to evaluate some of the remaining terms & see the change of period in the terms of the given sequence then you can find the value of the given term asked to compute. This can be illustrated by an example:

Ex.1. Let f1=f2=1, f3=-1,and also fn=fn-1*fn-3 . Find f1964 .
Sol:-  f1=f2=1, f3=f4=f5=-1,f6=1,f7=-1,and so on.
We see that the sequence starts with 1,1,-1,-1,-1,1,-1,1,1,-1,-1,-1,1-1, etc.
So,See the periodicity & check that the sequence is repeated after 7 terms.
Since 1964=(7*280)+4.
So  f1964 is nothing but the term f4=-1.

My Observation-2:- In some cases, you can see the initial terms are given & you are to find the value of an function. So,here you are required to derive the recurrence relation first & then apply the previous method. This can also be illustrated by an example:
Ex.2. If  x+ 1/x = -1, then find the value of    x^99 + 1/(x^99) .
Sol:-  Here the recurrence relation is
   fn+1 =fn*f1 - fn-1 ,for n>0,where
        fn = x^n + 1/(x^n) .
So,we are to find f99.
Now see that   f1=-1,  f3=f5=f7= . . .  .=2,
so, hence we have  f99=2.