Question

Find the sum of the series 1/(x + 1) + 2/(x² + 1) + 2²/(x⁴ + 1) + - - - - - - + 2¹⁰⁰/(x²^¹⁰⁰ + 1)
(when x = 2)

a) 1 - 2¹⁰¹/(4¹⁰¹ - 1)
b) 1 - 2¹⁰⁰/(4¹⁰⁰ - 1)
c) 1 + 2¹⁰¹/(4¹⁰¹ - 1)
d) 1 + 2¹⁰⁰/(4¹⁰¹ - 1)

__________

→ Above series in the form latex:
$\dfrac{1}{x + 1} + \dfrac{2}{ {x}^{2} + 1}+\dfrac{ {2}^{2}}{ {x}^{4} + 1} + \: - \: - \: - \: - \: + \dfrac{2^{100} }{{{x}^{2} }^{100}}$

Don't spam.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given series is

$\rm \: \dfrac{1}{x + 1} + \dfrac{2}{ {x}^{2} + 1} + \dfrac{ {2}^{2} }{ {x}^{4} + 1} + - - - + \dfrac{ {2}^{100} }{ {x}^{200} + 1}$

Let assume that

$\rm \: S = \dfrac{1}{x + 1} + \dfrac{2}{ {x}^{2} + 1} + \dfrac{ {2}^{2} }{ {x}^{4} + 1} + - - + \dfrac{ {2}^{100} }{ {x}^{200} + 1}$

can be rewritten as

$\rm \: - S = - \dfrac{1}{x + 1} - \dfrac{2}{ {x}^{2} + 1} - \dfrac{ {2}^{2} }{ {x}^{4} + 1} - - - \dfrac{ {2}^{100} }{ {x}^{200} + 1}$

can be further rewritten as

$\rm \: \dfrac{1}{x - 1} - S =\dfrac{1}{x - 1} - \dfrac{1}{x + 1} - \dfrac{2}{ {x}^{2} + 1} - \dfrac{ {2}^{2} }{ {x}^{4} + 1} - - - \dfrac{ {2}^{100} }{ {x}^{200} + 1}$

$\rm \: \dfrac{1}{x - 1} - S =\bigg(\dfrac{1}{x - 1} - \dfrac{1}{x + 1}\bigg) - \dfrac{2}{ {x}^{2} + 1} - \dfrac{ {2}^{2} }{ {x}^{4} + 1} - - - \dfrac{ {2}^{100} }{ {x}^{200} + 1}$

$\rm \: \dfrac{1}{x - 1} - S =\bigg(\dfrac{x + 1 - x + 1}{(x - 1)(x + 1)} \bigg) - \dfrac{2}{ {x}^{2} + 1} - \dfrac{ {2}^{2} }{ {x}^{4} + 1} - - - \dfrac{ {2}^{100} }{ {x}^{200} + 1}$

$\rm \: \dfrac{1}{x - 1} - S =\dfrac{2}{ {x}^{2} - 1} - \dfrac{2}{ {x}^{2} + 1} - \dfrac{ {2}^{2} }{ {x}^{4} + 1} - - - \dfrac{ {2}^{100} }{ {x}^{200} + 1}$

$\rm \: \dfrac{1}{x - 1} - S =\dfrac{2( {x}^{2} + 1 - {x}^{2} + 1)}{({x}^{2} - 1)( {x}^{2} + 1)} - \dfrac{ {2}^{2} }{ {x}^{4} + 1} - - - \dfrac{ {2}^{100} }{ {x}^{200} + 1}$

$\rm \: \dfrac{1}{x - 1} - S =\dfrac{4}{{x}^{4} - 1} - \dfrac{ {2}^{2} }{ {x}^{4} + 1} - - - \dfrac{ {2}^{100} }{ {x}^{200} + 1}$

$\rm \: \dfrac{1}{x - 1} - S =\dfrac{ {2}^{2} }{{x}^{4} - 1} - \dfrac{ {2}^{2} }{ {x}^{4} + 1} - - - \dfrac{ {2}^{100} }{ {x}^{200} + 1}$

So, on proceeding like this, we get

$\rm \: \dfrac{1}{x - 1} - S =\dfrac{ {2}^{100} }{{x}^{200} - 1} - \dfrac{ {2}^{100} }{ {x}^{200} + 1}$

$\rm \: \dfrac{1}{x - 1} - S =\dfrac{ {2}^{100}( {x}^{200} + 1 - {x}^{200} + 1)}{({x}^{200} - 1)( {x}^{200} + 1)}$

$\rm \: \dfrac{1}{x - 1} - S =\dfrac{ {2}^{100}(2)}{{x}^{400} - 1}$

$\rm \: \dfrac{1}{x - 1} - S =\dfrac{ {2}^{101}}{{x}^{400} - 1}$

$\rm\implies \:\rm \:S \: = \: \dfrac{1}{x - 1} - \dfrac{ {2}^{101}}{{x}^{400} - 1}$

Now, On substituting x = 2, we get

$\:\rm \:S \: = \: \dfrac{1}{2 - 1} - \dfrac{ {2}^{101}}{{2}^{400} - 1}$

$\:\rm \:S \: = \: 1 - \dfrac{ {2}^{101}}{{2}^{400} - 1}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

SHORT CUT TRICK

Sum of the series of the form

$\sf\: \dfrac{1}{x + 1} + \dfrac{2}{ {x}^{2} + 1} + \dfrac{ {2}^{2} }{ {x}^{4} + 1} + - - - + \dfrac{ {2}^{n} }{ {x}^{2n} + 1} = \dfrac{1}{x - 1} - \dfrac{ {2}^{n + 1} }{ {x}^{4n} - 1}$

Answer 2

Answer:

1 - 2¹⁰¹/(4¹⁰¹ - 1)

b) 1 - 2¹⁰⁰/(4¹⁰⁰ - 1)

c) 1 + 2¹⁰¹/(4¹⁰¹ - 1)

d) 1 + 2¹⁰⁰/(4¹⁰¹ - 1)

Step-by-step explanation:

hi dear how are u

Can be friends

dear can u msz me here