Missing two digits

The number is equal to $(\frac{10}{99})^2=(0.101...)^2$ .

Writing this out like a standard decimal multiplication, it becomes clear that his is equal to $\frac{1}{100}+\frac{2}{10000}+...+\frac{n}{10^{2n}}+...$ .

When $n>=100$ , adding up the terms is not merely concatenating the two-digit numbers anymore, we also get a carry : the 1 of the 100 adds up to 99, which again gives a carry that changes 98 into 99. Ergo just the two-digit sequence $\boxed{98}$ is missing.

from here we know that $B_2=0.\blue{02}\red{03}\blue{04}...\red{97}\blue{99}0...=\dfrac{199}{99^2}$ $\Rightarrow 1.\blue{02}\red{03}\blue{04}...\red{97}\blue{99}0...=\dfrac{199}{99^2}+1=\dfrac{100^2}{99^2}$ $\Rightarrow 0.\red{01}\blue{02}\red{03}\blue{04}...\red{97}\blue{99}0...=\dfrac{100}{99\times 99}$ $\because\space 98 \space is \space not \space in \space \dfrac{100}{99\times 99} \space the \space answer \space is \space \boxed{98}$

Let a = 100/99. We can easily see that a= 1.0101010101... and a/100 = 0.01010101010101...

Let b= a/99 , which is the number Brian is working on.

99b = a

100b = a + b

b = a/100 + b/100

Replace b on the right hand side with a/100 + b/100, we get

b = a/100 + a/10000 + b/10000

If we repeat the step until reaching near infinity, we'll get..

b = a/100 + a/10000 + a/1000000 + a/100000000 + ....

If we write it in decimal, we can see that

b =

0.01010101010101... +

0.00010101010101... +

0.00000101010101... +

0.00000001010101... + ....

= 0.0102030405060708.....

However, there is a problem with Brian's assumption. The resulting sum of this equation in the 199th and 200th place after decimal cannot be assumed as a reset toward ..00.. , but rather, it's actually ..100... giving an extra 1 to 198th digit. Although the 197th and 198th digit is a sum result of ..01.. 99 times, it also receive an extra 1 from the sum of the next digit, and thus, it also becomes ..100.. It gives an extra 1 to 196th digit. Thus, although digit number 195th and 196th are supposedly 98, since it receives the extra 1 from the next adjacent digit, it becomes 99. The number becomes..

b=0.0102030405.....9495969799000102030405060708.....

Okay so, Brian has a number N=100/99². And without any basis he claims it to be another number S=0.0102....99000102...and so on. We have to confirm whether S=N or not.
S=1/100+2/100²+...99/100⁹⁹+1/100¹⁰¹...and so on =(1/100+2/100²+...99/100⁹⁹)×(1+1/100+1/100²+...).
This expression is written based on the repetitive nature of S. So, we have written S as product of 2 numbers, first is a finite AGP, second is an infinite GP. After calculating, S=(1-(100×99/100¹⁰⁰-1))×(100/99²) which not the same as N. Thus Brian was wrong.
Now all we left to find is the missing 2digit numbers. Let r=1/100. Then, N=r/(1-r)². Consider 2 numbers 0<a,b<1.
1/(1-a)=1+a+a²+a³+......... 1/(1-b)=1+b+b²+b³+........ Subtract the 2 equations,
(a-b)/((1-a)(1-b))=(a-b)+(a²-b²)+(a³-b³)+...... 1/(1-a)(1-b)=1+(a+b)+(a²+ab+b²)+(a³+a²b+ab²+b³)+..........
Put a=b=r, and we have 1/(1-r)²=1+2r+3r²+........
Here, there's one problem, we have put a=b=r after canceling factor (a-b) from both sides which implies a cannot be equal to b, thus we have violated that condition. In order to counter that problem, we can do this: take a=r,b=r+h, h>0. Thus, now there's no violation of any condition but we have a new variable h to take care of. For that, what we can do is take h to be a very small positive number. As we keep taking smaller and smaller values of h(ie, taking the limit of h from right), h will almost become 0, and disappear from the equation and we will have our result. It might seem like an invalid method for those who are not familiar with the concept of limits and have no knowledge in calculus. Moving on, now we can write N=r/(1-r)²=r(1+2r+3r²+...) =r+2r²+3r³+... Consider the terms 98/100⁹⁸+99/100⁹⁹+100/100¹⁰⁰ =98/100⁹⁸+99/100⁹⁹+1/100⁹⁹= 98/100⁹⁸+100/100⁹⁹ =99/100⁹⁸. So in the 98th position we will find 99. Thus 98 has disappeared from our number. Thus the answer is 98.

Use the generalized geometric series:

$\begin{aligned} \sum_{k=0}^\infty \binom{k+n}{k}q^k&=\frac{1}{(1-q)^{n+1}},&&&|q|&<1, &n&\in\mathbb{N}&(*) \end{aligned}$

Let Brian's number be $X$ . Then we may rewrite it as $X=\frac{100}{99^2}=\frac{100}{(100-1)^2}=\frac{10^{-2}}{(1-10^{-2})^2}\underset{(*)}{=}10^{-2}\sum_{k=0}^\infty \binom{k+1}{k}10^{-2k}=\sum_{k=1}^\infty k10^{-2k}=:\sum_{k=1}^\infty a_k\qquad\left|\begin{aligned} k'&:=k+1,\\ k'&\rightarrow k \end{aligned}\right.$ If we only consider the first $99$ elements, we have no overlaps. But when we add $a_{100}$ , we have an overlap: $\sum_{k=1}^{100} a_k=\sum_{k=1}^{99} a_k + 100\cdot 10^{-200}=0.01\ldots 97989\red{9}+0.00\cdots00000\red{1}=0.01\ldots 979900$ We notice the number $98$ has changed to $99$ : The number $98$ is now missing! However, we do not know yet if the remainder of the series might grow big enough to change even more of the first $198$ digits after the decimal. To prove it doesn't, we need to show $\left|\sum_{k=101}^\infty a_k\right|<10^{-198}$ Let's try to find an upper estimate for the remainder: $\begin{aligned} \left| \frac{a_{k+1}}{a_k} \right| &= \frac{k+1}{k}\cdot 10^{-2}\leq\frac{102}{101}\cdot 10^{-2} =: r<\frac{1}{2} &\text{for}&& k&\geq 101 &&&\Rightarrow &&&& |a_{k}|&\leq |a_{101}| r^{k-101} &\text{for}&& k&\geq 101 &(**) \end{aligned}$ With these estimates, we can manage the remainder: $0\leq\left|\sum_{k=101}^\infty a_k\right| \underset{\Delta-\text{Ineq.}}{\leq}\sum_{k=101}^\infty |a_k| \leq |a_{101}|\sum_{k?101}^\infty r^{k-101}=101\cdot 10^{-202}\cdot\frac{1}{1-r}\underset{(**)}{<}101\cdot 10^{-202}\cdot 2<10^{-198}$ The remainder of the series leaves the first $198$ digits unchanged, so the only missing two-digit number is indeed $\boxed{98}$