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
105 Points
9 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 628 Points
3 years 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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