The value of 1√5+√3+1√7+√5+1√9+√7+⋯⋯ upto 100 terms is
A
100√203+√3
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B
100√203−√3
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C
200√218+√12
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D
200√218−√12
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Solution
The correct option is A100√203+√3 Let Sn=1√5+√3+1√7+√5+1√9+√7+⋯⋯+Tn So Tn=1√2n+3+√2n+1,n∈N ∴Tn=√2n+3−√2n+12=12[√2n+3−√2n+1] So, Sn=T1+T2+T3+⋯⋯+Tn=12[√5−√3]+12[√7−√5]+12[√9−√7]+⋮⋮+12[√2n+3−√2n+1]Sn=12[√2n+3−√3]⇒Sn=n√2n+3+√3∴S100=100√203+√3