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Question

If a=31/223+1 and f(n)=nr=1(1)r1 nCr1anr for all n3, then the value of f(2007)+f(2008) is

A
36
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B
39
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C
312
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D
315
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Solution

The correct option is B 39
f(n)=nr=1(1)r1 nCr1anr=1anr=1(1)r1 nCr1anr+1=1a[ nC0an nC1an1++(1)n1 nCn1a1]=1a[(a1)n(1)nCn]=1a[3n/223(1)n]f(n)=3n/223(1)n(31/223+1)
Now,
f(2007)+f(2008)=32007/223+32008/22331/223+1=39+32008/22331/223+1=39[1+31/22331/223+1]=39

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