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 - <p>25/101<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 2 - <p>99/400<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 3 - <p>101/408<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 4 - <p>101/404</p>
6 Views|Posted 5 months ago
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5 months ago
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
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Maths Ncert Solutions class 12th 2026
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