If the value of (1 + 2/3 + 6/3² + 10/3³ + ...upto ∞) / (log?.??(1/3 + 1/3² + 1/3³ + ...upto ∞)) is l, then l² is equal to……….
If the value of (1 + 2/3 + 6/3² + 10/3³ + ...upto ∞) / (log?.??(1/3 + 1/3² + 1/3³ + ...upto ∞)) is l, then l² is equal to……….
Asked by Shiksha User
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1 Answer
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The expression inside log is a GP sum: (1/3)/ (1-1/3) = 1/2.
log? (1/2) = 1/2.
The series S = 1+2/3+6/3²+.
S - S/3 = 1 + 4/3² + 4/3³ + . = 1 + (4/9)/ (1-1/3) = 1+2/3=5/3.
(2/3)S = 5/3 ⇒ S=5/2.
l = (5/2)/ (1/2)=5. l²=25.
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