Let a=(41/401−1) and for each n≥2, let bn=nC1+nC2⋅a+nC3⋅a2+⋯ – InterviewSolution

1.	Let a=(41/401−1) and for each n≥2, let bn=nC1+nC2⋅a+nC3⋅a2+⋯+nCn⋅an−1. Then the value of (b2006−b2005) is
Answer» Let $a = (4^{1 / 401} - 1)$ and for each $n \geq 2,$ let $b_{n} =^{n} C_{1} +^{n} C_{2} \cdot a +^{n} C_{3} \cdot a^{2} + \dots +^{n} C_{n} \cdot a^{n - 1} .$ Then the value of $(b_{2006} - b_{2005})$ is

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