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Questions

If $x^{n} = a_{0} + a_{1} (1 + x) + a_{2} (1 + x)^{2} + \dots$ …….. $+ a_{n} (1 + x)^{n}$
$\begin{array}{l} = b_{0} + b_{1} (1 - x) + b_{2} (1 - x)^{2} + \dots + b_{n} (1 - x)^{n} \end{array}$
then for $n = 201, (a_{101}, b_{101})$ is equal to:

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a

−201C101,−201C101

b

201C101,−201C101

c

−201C101,201C101

d

201C101,201C101

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detailed solution

Correct option is B

xn=[(1+x)−1]n=[1−(1−x)]n=∑k=0n nCk(1+x)n−k(−1)k=∑k=0n nCk(−1)k(1−x)k ∴ ak= coefficient of (1+x)k in∑k=0n nCk(1+x)n−k(−1)k=(−1)n−knCn−k=(−1)n−knCkand bk=nCk(−1)kFor n=201,k=101, we get a101,b101= 201C101,−201C101

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