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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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