If a=31/223+1 and f(n)=n∑r=1(−1)r−1nCr−1⋅an−r for all n≥3, 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 B39 f(n)=n∑r=1(−1)r−1nCr−1⋅an−r=1an∑r=1(−1)r−1nCr−1⋅an−r+1=1a[nC0an−nC1an−1+⋯+(−1)n−1nCn−1a1]=1a[(a−1)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