#progression
#progressionpyq's
#class11maths
#series
#arithmaticprogression
#specialseries
1/(3^2-1)+1/(5^2-1)+1/(7^2-1)+⋯+1/((201)^2-1) is equal to
(a) 101/404 (b) 25/101
(c) 101/408 (d) 99/400
Ans: b
Sol.
T_r=1/((2r+1)^2-1)=1/4r(r+1)
=1/4 (1/r-1/(r+1))
Now, ∑_(r=1)^100▒〖T_r=1/4〗 (1-1/101)=1/4×100/101=25/101
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