If xn=a0+a1(1+x)+a2(1+x)2+………..+an(1+x)n=b0+b1(1−x)+b2(1−x)2+…+bn(1−x)n then for n=201,a101,b101 is equal to:

A
$(-^{201} C_{101}, -^{201} C_{101})$
B
$(^{201} C_{101}, -^{201} C_{101})$
C
$(-^{201} C_{101},^{201} C_{101})$
D
$(^{201} C_{101},^{201} C_{101})$

Solution:

$\begin{array}{l} x^{n} = [(1 + x) - 1]^{n} = [1 - (1 - x)]^{n} \\ = \sum_{k = 0}^{n}^{n} C_{k} (1 + x)^{n - k} (- 1)^{k} \\ = \sum_{k = 0}^{n}^{n} C_{k} (- 1)^{k} (1 - x)^{k} \end{array}$

$∴$ $a_{k} =$ coefficient of $(1 + x)^{k}$ in

$\sum_{k = 0}^{n}^{n} C_{k} (1 + x)^{n - k} (- 1)^{k} = (- 1)^{n - k^{n}} C_{n - k} = (- 1)^{n - k^{n}} C_{k}$

and $b_{k} =^{n} C_{k} (- 1)^{k}$

For

$\begin{array}{l} n = 201, k = 101, \end{array}$

we get $(a_{101}, b_{101}) = (^{201} C_{101}, -^{201} C_{101})$

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:

Solution:

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