A sequence of real numbers (x_n) is defined recursively as follows x_0=a and x_1=b are positive real numbers, and x_(n+2)=(1+x_(n+1))/(x_n) for n=0,1,2,... Find the value of x_2012, in terms of a and b.
+23251 x0 = a
x1 = b
xn + 2 = ( 1 + xn+ 1 ) / xn
x2012 = ?
Let's write out a few terms:
x0 = a
x1 = b
x2 = ( 1 + b ) / a = ( b + 1 ) / a
x3 = [ 1 + ( b + 1 ) / a ] / b
Multiply this term by a / a and simplify:
x3 = [ 1 + ( b + 1 ) / a ] / b · a / a ---> x3 = [ a + ( b + 1 ) ] / ( ab )
x3 = ( a + b + 1 ) / ( ab )
x4 = [ 1 + ( a + b + 1 ) / ( ab ) ] / ( b + 1 ) / a
Multiply this term by (ab) / (ab) and simplify:
x4 = [ 1 + ( a + b + 1 ) / ( ab ) ] / ( b + 1 ) / a · ( ab ) / ( ab ) ---> [ ab + a + b + 1 ] / [ b( b + 1) ]
Factor the numerator: ab + a + b + 1 = a(b + 1) + (b + 1) = (a + 1)(b + 1)
x4 = [ (a + 1)(b + 1) ] / [ b( b + 1) ]
x4 = (a + 1) / b
x5 = [ 1 + (a + 1)/b ] / [ (a + b + 1)/(ab)
Multiply this term by (ab)/(ab) and simplify:
x5 = [ ab + a(a + 1) ] / [ (a + b + 1) ] = [ a(b + a + 1 ] / (a + b + 1) ]
x5 = a Which is just x0!
x6 = [ 1 + a ] / [ (a + 1) / b ]
Multiply this term by b/b and simplify:
x6 = [ b(1 + a) ] [ (a + 1) ]
x6 = b Which is just x1!
This shows that it repeats every 5 terms.
To find out the value of term #2012, divide 2012 by 5 and note the remainder.
2012 / 5 = 402 remainder 2.
Thus term #2012 equals terms #2 ---> x2012 = x2 = (b + 1) / a
