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Let a=41/401−1 and for each n≥2, let bn=nC1+nC2⋅a+nC3⋅a2+⋯+nCn⋅an−1. If the value of b2006−b2005 is 4k, where k∈N, then the value of k is

Answer» Let a=41/4011 and for each n2, let bn=nC1+nC2a+nC3a2++nCnan1. If the value of b2006b2005 is 4k, where kN, then the value of k is


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