1.

Let a=31/203+1 and f(n)= nC0an−1− nC1an−2+ nC2an−3−⋯+(−1)n−1 nCn−1a0 where n≥3. If f(2030)+f(2031)=3x(ya−1), then the value of xy is

Answer»

Let a=31/203+1 and f(n)= nC0an1 nC1an2+ nC2an3+(1)n1 nCn1a0 where n3. If f(2030)+f(2031)=3x(ya1), then the value of xy is



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