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the sum to infinite terms of the series tan -1 (1/3) + tan -1 (2/9) +...................+ tan -1 ( 2 n-1 / 1+2 2n-1 )....is?

the sum to infinite terms of the series tan-1(1/3) + tan-1 (2/9) +...................+ tan-1( 2n-1/ 1+22n-1)....is?

Grade:12

2 Answers

Abhishek Singh
93 Points
6 years ago
the  general term ( r th term) for series is Tan-1 ( 2r-1/ 1+22r-1) = Tan-1 ( 2r-1/ 1+ 2r-1 * 2r
 which equals  tan-1 2 – tan-1 2r-1.
Now series sum =   tan-1 2 – tan-1 2r-1.  and r runs from 1 to n
so answer = tan-12- tan-120 = tan-12- pi/4
 
Ithink you know that   f(r+1)-f(r)      where r runs from m to n is equal to f(m+1)-f(m)
that is simply we have to dump the upper limit in the first function and the lower to second one.
 
           
Kushagra Madhukar
askIITians Faculty 629 Points
one year ago
Hello student
 
In the given question the rth term can be written as
Tr = tan-1(2r-1/1 + 22r-1
Now the neumerator can be written as, 2r-1 = 2r-1 x 1 = 2r-1 x (2 -1) = 2r – 2r-1
and the denominator can be written as, 1 + 22r-1 = 1 + 2r.2r-1
let, 2r = x; 2r-1 = y
hence,
Tr = tan-1(x – y/1 + x.y) 
which is an inverse trigonometric identity and can be written as
Tr = tan-1(x) – tan-1(y) = tan-1(2r) – tan-1(2r-1)
 
The given sum is the summation of Tr from r = 0 to r = n
∑Tr = ∑(tan-1(2r) – tan-1(2r-1)) 
on simplifying we get
∑Tr = tan-12n – tan-120 = tan-12n – tan-11 = tan-12n – π/4
 
Hope it helps
Regards,
Kushagra

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