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Question : 62 of 160
Marks:
+1,
-0
Solution:
∵∴x=1+‌×‌+‌(‌)2 +‌(‌)3+... ∴‌‌(1−α)−p∕q +‌α3+... =1+‌(‌)+‌(‌)2 +‌(‌)3+... On comparing Eqs. (i) and (ii), we get
p=3,p+q=7,p+2q=11 and
‌‌‌=‌⇒q=4 and
α=‌=‌ So, let
X=(1−α)−p∕q=(1−‌)−3∕4 =(‌)−3∕4=(3)3∕4 ∴‌‌x4=33=27
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