11can i get the solution?
if the sequence {an} ,satisfies the recurrence ,
an+1 = 3an -2an-1 for n>=2
a0 = 2
a1 = 3
Find a2007
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12 Answers
106see what bhatt sir has written
an+1 = 3an - 2an-1
=> (an+1 - an) = 2(an-an-1))
If we denote an-an-1 as bn then, we get
bn+1 = 2bn = 22bn-1 = ....2n-1b2
bn = 2n-2b2
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.
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b2 = b2
adding vertically, we get
an+1 - a1 = b2 (1+ 2+ 22 + ...+ 2n-1) = b2(2n-1)
taking n = 2006, we get
a2007 - a1= (a2 - a1)*(22006 -1)
=> a2007 -3 = 2*22006 - 2
=> a2007 = 22007 +1
one mistake in the question is that the recursive relation is valid for n≥1 and not n≥2 as given
62To see if you have understood this question try to solve a very small variation (dont know if this should be called a variation at the first place :P)
2an+1= 5an -3an-1 for n>=1
find the general value of an
in terms of a1 and a2
11There was one more series qn.
The sequence {an) is an increasing integers with only positive integers.
an+2 = an+1 + anfor n > 1
a7 = 120
find a8
1That ' s the Fibonacci Sequence , except for the fact that the terms is different . However , I ' ll rather not spoil the fun for others to solve this problem by giving a rigorous solution .
1That ' s the Fibonacci Sequence , except for the fact that the terms is different . However , I ' ll rather not spoil the fun for others to solve this problem by giving a rigorous solution .