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The question says to find the value of $x$ if, $$2^x\Bigl(2^x-1\Bigl) + 2^{x-1}\Bigl(2^{x-1} -1 \Bigl) + .... + 2^{x-99}\Bigl(2^{x-99} - 1 \Bigl)= 0$$
My approach:
I rewrote the expression as, $$2^x\Bigl(2^x-1\Bigl) + \frac{2^x}{2}\Bigl(\frac{2^x}{2} -1 \Bigl) + .... + \frac{2^x}{2^{99}} \Bigl(\frac{2^x}{2^{99}} - 1 \Bigl)= 0$$ I then took $\bigl(2^x\bigl)$ common and wrote it as, $$2^x \Biggl[ \Bigl(2^x - 1\Bigl) + \frac{1}{2^1}\Bigl(2^x -2^1\Bigl) + \frac{1}{2^2}\Bigl(2^x - 2^2\Bigl) + \;\ldots + \frac{1}{2^{99}} \Bigl(2^x - 2^{99}\Bigl)\Biggl] = 0$$ After further simplification I got, $$\frac{2^x}{2^{99}} \Biggl[ \Bigl(2^x\cdot2^{99} - 2^{99}\Bigl) + \Bigl(2^x \cdot 2^{98} - 2^{99}\Bigl) + \ldots + \bigl(2^x -2^{99}\bigl)\Biggl] = 0$$ Taking $-2^{99}$ common I got, $$-2^x \Biggl[ \Bigl( 2^{x+99} + 2^{x+98} + \ldots + 2^{x+2} + 2^{x+1} + 2^x \Bigl)\Biggl]= 0$$ Now the inside can be expressed as $$\sum ^ {n= 99} _{n=1} a_n$$ Where $a_n$ are the terms of the GP.
Thus we can see that either $$-2^x= 0$$ Or, $$\sum ^ {n= 99} _{n=1} a_n = 0$$ Since the first condition is not poossible, thus, $$\sum ^ {n= 99} _{n=1} a_n = 0$$ So, $$2^{x + 99} \Biggl(\frac{1-\frac{1}{2^{100}}}{1-\frac{1}{2}} \Biggl) = 0$$ Either way once I solve this, I am not getting an answer that is even in the options. The answers are all in the form of logarithmic expressions.

Any help would be appreciated. We have to find the value of $x$.

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  • $\begingroup$ I understand now, how to reach at the answer, but can someone also explain what I have done wrong? $\endgroup$ Commented Jul 28, 2018 at 12:31
  • $\begingroup$ "can someone also explain what I have done wrong?" At the step "Taking $−2^{99}$ common I got", you forgot all the negative terms. $\endgroup$
    – Did
    Commented Jul 28, 2018 at 13:19

5 Answers 5

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$$2^x\Bigl(2^x-1\Bigl) + 2^{x-1}\Bigl(2^{x-1} -1 \Bigl) + .... + 2^{x-99}\Bigl(2^{x-99} - 1 \Bigl)= $$

$$2^{2x}-2^x + 2^{2x-2}-2^{x-1} + .... + 2^{2x-198}-2^{x-99}= $$ $$(2^{2x}+ 2^{2x-2} .... + 2^{2x-198})-(2^x +2^{x-1} + .... + 2^{x-99})= $$ $$2^{2x-198}(2^{198}+ 2^{196} .... + 2^{0})-2^{x-99}(2^{99} +2^{98} + .... + 2^{0})= $$ $$2^{2x-198}{2^{200}-1\over 2^2-1} -2^{x-99}{2^{100}-1\over 2-1}=0 $$

If we cancel this with $2^{x-99}(2^{100}-1)$ we get

$$2^{x-99}{2^{100}+1\over 3} -1=0 $$ So $$2^{x-99} = {3\over 2^{100}+1}$$ and thus $$x= 99+\log_{2} {3\over 2^{100}+1}$$

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\begin{align} 2^x\Bigl(2^x-1\Bigl) + 2^{x-1}\Bigl(2^{x-1} -1 \Bigl) + .... + 2^{x-99}\Bigl(2^{x-99} - 1 \Bigl) &= 0 \\ \sum_{i=0}^{99}2^{x-i}\left(2^{x-i}-1\right) &= 0 \\ \sum_{i=0}^{99}\left[\left(2^{x-i}\right)^2-2^{x-i}\right] &= 0 \\ \sum_{i=0}^{99}\left(2^{x-i}\right)^2-\sum_{i=0}^{99}2^{x-i} &= 0 \\ \sum_{i=0}^{99}2^{2x-2i}&=\sum_{i=0}^{99}2^{x-i} \\ \dfrac{2^{2x}}{\sum_{i=0}^{99}2^{2i}}&=\dfrac{2^{x}}{\sum_{i=0}^{99}2^{i}} \\ 2^x&=\dfrac{\sum_{i=0}^{99}2^{2i}}{\sum_{i=0}^{99}2^{i}} \\ &= \dfrac{4^{100}-1}{4-1}\cdot\dfrac{2-1}{2^{100}-1} \\ &= \dfrac{4^{100}-1}{3\left(2^{100}-1\right)}\\ x &= \log_2\left(\dfrac{4^{100}-1}{3\left(2^{100}-1\right)}\right) \end{align}

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From scratch, the quantity you are looking at is $$\sum_{k=0}^{99} 2^{2(x-k)}-2^{x-k}=4^x\sum_{k=0}^{99}4^{-k}-2^x\sum_{k=0}^{99}2^{-k}=4^x\cdot\frac{4^{-100}-1}{-\frac34}-2^x\frac{2^{-100}-1}{-\frac12}=\\=2^x\left(2^x\cdot \frac{4-4^{-99}}{3}-2+2^{-99}\right)$$

That quantity is $0$ if and ony if $$2^x=\frac{3\cdot 2}4\cdot\frac{1-2^{-100}}{1-4^{-100}}=\frac{3}{2(1+2^{-100})}\\ x=\frac{\ln 3-\ln(1+2^{-100})}{\ln 2}-1$$

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Here, you forgot red:

$$2^x \Biggl[ \Bigl(2^x - 1\Bigl) + \frac{1}{2^1\cdot \color{red}{2}}\Bigl(2^x -2^1\Bigl) + \frac{1}{2^2\color{red}{2^2}}\Bigl(2^x - 2^2\Bigl) + \;\ldots + \frac{1}{2^{99}\color{red}{2^{99}}} \Bigl(2^x - 2^{99}\Bigl)\Biggl] = 0$$

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  • $\begingroup$ But I have done the same thing in my answer, i simply wrote it as $\Bigl (2^x -1\bigl) + \frac {1}{2^1}\Bigl (\frac{2^x}{2} - 1\Bigl).... $. I simplified that expression in a later step. $\endgroup$ Commented Jul 28, 2018 at 12:48
  • $\begingroup$ This is what I see and I think you didn't take it in consideration in later steps. $\endgroup$
    – nonuser
    Commented Jul 28, 2018 at 13:52
  • $\begingroup$ Alright. I will check my solution again. $\endgroup$ Commented Jul 28, 2018 at 14:33
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Hint.

$$ 2^{2x}\sum_{k=0}^{99}2^{-2k} =2^x\sum_{k=0}^{99}2^{-k} $$

hence

$$ 2^x = \frac{\sum_{k=0}^{99}2^{-k}}{\sum_{k=0}^{99}2^{-2k}} = 1.5\to x = 0.5849625007211562 $$

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