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 = 101,a50,b50 equals:

A
$(-^{101} C_{50},^{101} C_{50})$
B
$(^{101} C_{50}, -^{101} C_{50})$
C
$(-^{101} C_{50}, -^{101} C_{50})$
D
$(^{101} C_{50},^{101} C_{50})$

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 $n = 101, k = 51$ we get

$(a_{51}, b_{51}) = (^{101} C_{51}, -^{101} C_{51})$

$x^{n} = a_{0} + a_{1} (1 + x) + a_{2} (1 + x)^{2} + \dots + a_{n}$ $(1 + x)^{n}$ $= b_{0} + b_{1} (1 - x) + b_{2} (1 - x)^{2} + \dots + b_{n} (1 - x)^{n}$ then for $n = 101,$ $(a_{50}, b_{50})$ equals:

Solution:

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