1/(3²-1) + 1/(5²-1) + 1/(7²-1) + ... + 1/((201)²-1) is equal to:
1/(3²-1) + 1/(5²-1) + 1/(7²-1) + ... + 1/((201)²-1) is equal to:
Option 1 -
25/101
Option 2 -
99/400
Option 3 -
101/408
Option 4 -
101/404
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1 Answer
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Correct Option - 1
Detailed Solution:Consider the series:
1/ (3²-1) + 1/ (5²-1) + 1/ (7²-1) + . + 1/ (201)²-1)The general term T? can be written for the r-th term starting with r=1 for 3, r=2 for 5.
T? = 1/ (2r+1)² - 1) = 1/ (2r+1-1) (2r+1+1) = 1/ (2r * (2r+2) = 1/4 * 1/ (r (r+1)
T? = 1/4 * (1/r - 1/ (r+1)The sum of the first n terms is:
S? = Σ T? = 1/4 * Σ (1/r - 1/ (r+1) from r=1 to n
S? = 1/4 * [ (1 - 1/2) + (1/2 - 1/3) + . + (1/n - 1/ (n+1) ]
S? = 1/4 * (1 - 1/ (n+1)The last term is (201)²-1, so 2r+1 = 201, which gives r = 100. So, n=100.
S? = 1/4 * (1 - 1/101) = 1/4 * (100/101) = 25/101.
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