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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?

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5 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 2r  – tan-1 2r-1.Now series sum =  ∑ tan-1 2r  – tan-1 2r-1.  and r runs from 1 to nso answer = tan-12n - tan-120 = tan-12n - 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.
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5 years ago 612 Points
```							Hello student In the given question the rth term can be written asTr = 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-1and the denominator can be written as, 1 + 22r-1 = 1 + 2r.2r-1let, 2r = x; 2r-1 = yhence,Tr = tan-1(x – y/1 + x.y) which is an inverse trigonometric identity and can be written asTr = 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 helpsRegards,Kushagra
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7 months ago
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