f(n)= nC0an−1− nC1an−2+ nC2an−3+…+ (−1)n−1 nCn−1a0
=1a( nC0an− nC1an−1+ nC2an−2+…+ (−1)n−1 nCn−1a)
=1a((a−1)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)=32008/223−131/223+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