1.

sum of the series upto infinity from 0 to infinitytan-1(1/1+1+12)+tan-1(1/1+2+22)+tan-1(1/1+3+32)....................

Answer»

The general term of the expression is

tr.  = tan^-1(1/(1+r+r^^2))

=>t=tan^-1(r+1)-tan^-1(r)

So inserting r=1 to infinity and summing up we get the required sum as

S =t1+t2+t3+.......tinfinity

=tan^-1(infinity)-tan^-1(1)

=pi/2-pi/4=pi/4



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