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Controversy in Mathematics Books
This is the "cumulative multiplication" of the general term of the sequence, which is suitable for the recurrence relation of quotient type. Similarly, if the differential recurrence relation is given, the "accumulation method" is considered.

An = (2n-3)/(2n+1) × a (n-1), then: an/[a (n-1)] = (2n-3)/(2n+1).

a2/a 1= 1/5

a3/a2=3/7

a4/a3=5/9

a5/a4=7/ 1 1

……

an/a(n- 1)=[2n-3]/[2n+ 1]

Multiply all the above formulas to get:

An/a1= 3/[(2n-1) (2n+1)], substituting a 1= 1/3, with: an= 1/[4n? - 1]