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Question
Find the sum of n terms of the series 1√1+√3+1√3+√5+1√5+√7
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Solution
The required sum is1√1+√3+1√3+√5+1√5+√7+....+1√2n−1+√2n+1
=√3−√1−(√1)2+(√3)2+√5−√3−(√3)2+(√5)2+....+√2n+1−√2n−1−(√2n−1)2+(√2n+1)2
=12{(√3−√1)+(√5−√3)+......+(√2n+1−√2n−1)}
=12{√2n+1−1}
1√1+√3+1√3+√5+1√5+√7+....+1√2n−1+√2n+1
=√3−√1−(√1)2+(√3)2+√5−√3−(√3)2+(√5)2+....+√2n+1−√2n−1−(√2n−1)2+(√2n+1)2
=12{(√3−√1)+(√5−√3)+......+(√2n+1−√2n−1)}
=12{√2n+1−1}
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