$$xn=a0+a1(1+x)+a2(1+x)2+$$…$$+an(1+x)n=b0+b1(1$$−$$x)+b2(1$$−$$x)2+$$…$$+bn(1$$−$$x)n$$ then for n $$=$$ $$101$$,$$a50$$,$$b50$$ equals:

Correct option is 2101C50,−101C50

Questions

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

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−101C50,101C50

101C50,−101C50

−101C50,−101C50

101C50,101C50

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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=101,k=51 we geta51,b51= 101C51,−101C51

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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:

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