Maths-
General
Easy

Question

open square brackets 1 plus open parentheses 1 half plus 1 third close parentheses times 1 fourth plus open parentheses 1 fourth plus 1 fifth close parentheses times 1 over 4 squared plus open parentheses 1 over 6 plus 1 over 7 close parentheses times 1 over 4 cubed plus horizontal ellipsis horizontal ellipsis straight infinity close square brackets equals

  1. log square root of 12
  2. log square root of 11
  3. log 12
  4. log 11

hintHint:

In this question, we have given open square brackets 1 plus open parentheses 1 half plus 1 third close parentheses times 1 fourth plus open parentheses 1 fourth plus 1 fifth close parentheses times 1 over 4 squared plus open parentheses 1 over 6 plus 1 over 7 close parentheses times 1 over 4 cubed plus horizontal ellipsis horizontal ellipsis straight infinity close square brackets.  We have to find the sum of this series, for that open the bracket and find the sum.

The correct answer is: log square root of 12


    Here we have to find the sum of the series.
    Firstly, the given series is
    open square brackets 1 plus open parentheses 1 half plus 1 third close parentheses times 1 fourth plus open parentheses 1 fourth plus 1 fifth close parentheses times 1 over 4 squared plus open parentheses 1 over 6 plus 1 over 7 close parentheses times 1 over 4 cubed plus horizontal ellipsis horizontal ellipsis straight infinity close square brackets
    S = 1 plus 1 half x 1 fourth plus 1 third x 1 fourth plus 1 fourth x 1 over 4 squared plus 1 fifth x 1 over 4 squared plus 1 over 6 x 1 over 4 cubed......…..
    We can write
    S = 1 half x 1 fourth plus 1 fourth x 1 over 4 squared plus 1 over 6 x 1 over 4 cubed...... plus 1 plus 1 third x 1 fourth plus 1 fifth x 1 over 4 squared.....
    We know that,
    Log ( 1 + x ) = x minus x squared over 2 space plus space x cubed over 3 minus x to the power of 4 over 4 plus..
    Log (1 - x) = negative x minus x squared over 2 space minus space x cubed over 2 minus x to the power of 4 over 2 minus..- …
    Log ( 1 – x2) = -2 [x squared over 2 plus x to the power of 4 over 4 minus...]              ------(1)
    Now ,
    Log ( fraction numerator 1 space plus space x space over denominator 1 minus x space end fraction) = 2 [x plus x cubed over 3 plus x to the power of 5 over 5 plus...]
                             =2 x[ 1 plus x squared over 3 plus x to the power of 4 over 5 plus...]     ----(2)
    Replacing eq1 and eq2 in S, here x = ½
    S = -1/2 log ( 1 – ¼ ) + 1/( 2 x ½) log ( 1 + ½ / 1 – ½ )
    S = -½ log ( ¾ ) + log ( 3 )
    S = -½ log ( 3 ) + ½ log ( 4) + log ( 3 )
    S = ½ log ( 3 ) + ½ log ( 4)
    S = ½ log ( 12 )
    S = log √12
    Therefore, the correct answer is log log√12

    In this question, we have to find the sum of the series. Separate the series in even and odd series, and replace with Log ( 1 – x2) and Log ( 1 + x / 1-x ) .

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