1sorry..
its
tan-1[(k+1)/4] - tan-1[(k-1)/4]
12 Answers
11/2k2 = 1/4 [{(k+1) - (k-1)} / {1 + (k+1)(k-1)}]
so.. tan-1 (..) = tan-1[1/4 {(k+1)-(k-1)}
1afetr every 4 terms the terms get cancelled ...
only tan-1 0 remains...
so is tanθ= 0 ??
341tan-1(1/2k2) = tan-1(2/4k2) = tan-1[(2k+1)-(2k-1)/1+(2k+1)(2k-1)]
= tan-1 (2k+1) - tan-1(2k-1)
So this is a telescopic sum that sums to tan-1 (2n+1) - π/4
and when n→∞, the sum is π/2-π/4 = π/4
33[2]
ye to mujhe pata hai... par agar kisi ko pata nahi hai ki 1/2k2 ko aise likhna hai to wo kya karega...
subah 4 baje sapna aega kya ki 1 ko 1 nahi (5-4)(sin2x+cos2x) likna hai.. [3]
62phew.. this was the last trick I could have found out myself...
good work sky and as always prophet :)
341I was trying to add this problem as a new post. Its not working though.
The problem is very much on similar lines as this one.
Find Lt n→∞ 1Σn tan-12r/(2r4+1)
62\\tan^{-1}(2r^2+2r+1)-tan^{-1}(2r^2-2r+1)=\frac{(2r^2+2r+1)-(2r^2-2r+1)}{(2r^2+1)^2-(2r)^2-1} \\tan^{-1}(2r^2+2r+1)-tan^{-1}(2r^2-2r+1)=\frac{4r}{(2r^2+1)^2-(2r)^2-1} \\tan^{-1}(2r^2+2r+1)-tan^{-1}(2r^2-2r+1)=\frac{4r}{4r^4+2} \\tan^{-1}(2r^2+2r+1)-tan^{-1}(2r^2-2r+1)=\frac{2r}{2r^4+1}
Not a very simple soln to be fair..
but now i guess someone can finish this off [1]