Solveeit Logo
Login

Question

Question: The sum of the series $\frac{1}{x+1}+\frac{2}{x^2+1}+\frac{2^2}{x^4+1}+....+\frac{2^{100}}{x^{2^{100...

The sum of the series 1x+1+2x2+1+22x4+1+....+2100x2100+1\frac{1}{x+1}+\frac{2}{x^2+1}+\frac{2^2}{x^4+1}+....+\frac{2^{100}}{x^{2^{100}}+1} when x = 2 is: [JEE Main-2021(August)]

A

1+21014101−1\frac{2^{101}}{4^{101}-1}

B

1+21004101−1\frac{2^{100}}{4^{101}-1}

C

1-21004100−1\frac{2^{100}}{4^{100}-1}

D

1-210142100−1\frac{2^{101}}{4^{2^{100}}-1}

Answer

1-210142100−1\frac{2^{101}}{4^{2^{100}}-1}

Explanation

Solution

The given series is S=1x+1+2x2+1+22x4+1+....+2100x2100+1S = \frac{1}{x+1}+\frac{2}{x^2+1}+\frac{2^2}{x^4+1}+....+\frac{2^{100}}{x^{2^{100}}+1}. This can be written in summation notation as S=∑k=01002kx2k+1S = \sum_{k=0}^{100} \frac{2^k}{x^{2^k}+1}.

We use the algebraic identity: 1y−1−1y+1=(y+1)−(y−1)(y−1)(y+1)=2y2−1\frac{1}{y-1} - \frac{1}{y+1} = \frac{(y+1)-(y-1)}{(y-1)(y+1)} = \frac{2}{y^2-1}. Rearranging this identity to isolate the term with (y+1)(y+1) in the denominator, we get: 1y+1=1y−1−2y2−1\frac{1}{y+1} = \frac{1}{y-1} - \frac{2}{y^2-1}.

Let y=x2ky = x^{2^k}. Then y2=(x2k)2=x2⋅2k=x2k+1y^2 = (x^{2^k})^2 = x^{2 \cdot 2^k} = x^{2^{k+1}}. Substituting y=x2ky = x^{2^k} into the rearranged identity: 1x2k+1=1x2k−1−2x2k+1−1\frac{1}{x^{2^k}+1} = \frac{1}{x^{2^k}-1} - \frac{2}{x^{2^{k+1}}-1}.

Now, multiply both sides by 2k2^k to match the numerator of the terms in our series: 2kx2k+1=2kx2k−1−2k⋅2x2k+1−1=2kx2k−1−2k+1x2k+1−1\frac{2^k}{x^{2^k}+1} = \frac{2^k}{x^{2^k}-1} - \frac{2^k \cdot 2}{x^{2^{k+1}}-1} = \frac{2^k}{x^{2^k}-1} - \frac{2^{k+1}}{x^{2^{k+1}}-1}.

Now, we can write the sum SS using this expression: S=∑k=0100(2kx2k−1−2k+1x2k+1−1)S = \sum_{k=0}^{100} \left( \frac{2^k}{x^{2^k}-1} - \frac{2^{k+1}}{x^{2^{k+1}}-1} \right).

This is a telescoping sum. Let Ak=2kx2k−1A_k = \frac{2^k}{x^{2^k}-1}. The sum becomes ∑k=0100(Ak−Ak+1)\sum_{k=0}^{100} (A_k - A_{k+1}). The sum of such a series is A0−A101A_0 - A_{101}.

A0=20x20−1=1x1−1=1x−1A_0 = \frac{2^0}{x^{2^0}-1} = \frac{1}{x^1-1} = \frac{1}{x-1}. A101=2101x2101−1A_{101} = \frac{2^{101}}{x^{2^{101}}-1}.

So, the sum of the series is: S=1x−1−2101x2101−1S = \frac{1}{x-1} - \frac{2^{101}}{x^{2^{101}}-1}.

We are asked to find the sum when x=2x=2. Substitute x=2x=2 into the formula: S=12−1−210122101−1S = \frac{1}{2-1} - \frac{2^{101}}{2^{2^{101}}-1} S=1−210122101−1S = 1 - \frac{2^{101}}{2^{2^{101}}-1}.

Let's simplify the denominator of option (4): 42100−1=(22)2100−1=22⋅2100−1=22101−14^{2^{100}}-1 = (2^2)^{2^{100}}-1 = 2^{2 \cdot 2^{100}}-1 = 2^{2^{101}}-1. So, option (4) is 1−210122101−11-\frac{2^{101}}{2^{2^{101}}-1}.

This matches our derived sum.