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Question

Let a=31/223+1 and for all n3, let f(n)= nC0an1 nC1an2+ nC2an3+(1)n1a0. If the value of f(2007)+f(2008)=3k where kN, then the value of k is

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Solution

f(n)= nC0an1 nC1an2+ nC2an3++ (1)n1 nCn1a0

=1a( nC0an nC1an1+ nC2an2++ (1)n1 nCn1a)

=1a((a1)n(1)n nCn)

=1a((3n/223(1)n nCn))


f(x)=3n/223(1)n31/223+1

f(2007)=32007/223+131/223+1

f(2008)=32008/223131/223+1

f(2007)+f(2008)=32007/223+32008/22331/223+1

=39+39+1/22331/223+1

=39(31/223+131/223+1)=39

39=3k
Then, k=9

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