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Let a=(4^(1/401)-1) and let bn = nC1 ...

Let `a=(4^(1/401)-1)` and let `b_n = nC_1 + nC_2 . a+ nC_3. a^2+......+ nC_n .a^(n-1)`. Find the value of `(b_2006 – b_2005)`

Text Solution

Verified by Experts

The correct Answer is:
`2^(10)`
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