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If α,β0, and f(n)=αn+βn and |31+f(1)1+f(2)1+f(1)1+f(2)1+f(3)1+f(2)1+f(3)1+f(4)|=K(1α)2(1β)2(αβ)2, then K is equal to


( A ) αβ
( B ) 1
( C ) 1
( D ) 1αβ
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