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Question

If the expansion of 1(1ax)(1bx)=a0+a1x+a2x2++anxn+, then an is (where ab,|ax|,|bx|<1)
  1. anbnba
  2. bn+1an+1ba
  3. bnanba
  4. an+1bn+1ba


Solution

The correct option is B bn+1an+1ba
1(1ax)(1bx)=a0+a1x+a2x2+.....+anxn+....
but (1ax)1(1bx)1=(1+ax+a2x2+)(1+bx+b2x2+)
comparing the coefficient of xn in above equations
bn+abn1+a2bn2+....+an1b+an=an
an=bn+1an+1ba

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