Undoubtedly Rational

Relevant wiki: Arithmetic-Geometric Progression

$N = 1.02030405\ldots \\ \color{#D61F06}{N = 1 + 0.02 + 0.0003 + 0.000004 + 0.00000005 + \ldots} \\ \text{Multiply both side by `100'} \\ \color{#3D99F6}{100N = 100 + 2 + 0.03 + 0.0004 + 0.000005 + \ldots} \\ \text{Subtract it from N} \\ \color{#3D99F6}{100N = 100 + 2 + 0.03 + 0.0004 + 0.000005 + \ldots} \\ \color{#D61F06}{\text{ (-)N =} \quad\quad - 1 - 0.02 - 0.0003 - 0.000004-\ldots} \\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\\ \color{#3D99F6}{99N = 100 + 1 + 0.01 + 0.0001 + 0.000001 + 0.00000001 + \ldots}\\ \text{Now, combine all of those geometric progressions} \\ \color{#20A900}{99N = 100 + } \color{#3D99F6}{(1 + 0.1 + 0.0001 + 0.000001 + 0.00000001 + \ldots)} \\ \color{#3D99F6}{(1 + 0.1 + 0.0001 + 0.000001 + 0.00000001 + \ldots)} \text{ is in Geometric Progression, where a = 1, r = } \dfrac{1}{100} \\ \text{Therefore, as GP tends to infinity} \\ \color{#20A900}{99N = 100 + } \color{#3D99F6}{(\dfrac{a}{1-r})} \\ \color{#20A900}{99N = 100 + } \color{#3D99F6}{(\dfrac{1}{1-\dfrac{1}{100}})} \\ \color{#20A900}{99N = 100 + } \color{#3D99F6}{(\dfrac{1}{\dfrac{100 - 1}{100}})} \\ \color{#20A900}{99N = 100 + } \color{#3D99F6}{(\dfrac{100}{99})} \\ \color{#20A900}{99N = } \color{#3D99F6}{\dfrac{(100 \times 99) + 100}{99}} \\ \color{#20A900}{99N = } \color{#3D99F6}{\dfrac{(9900) + 100}{99}} \\ \color{#20A900}{99N = } \color{#3D99F6}{\dfrac{10,000}{99}} \\ \color{#3D99F6}{N} = \dfrac{10000}{9801} \\ \color{#D61F06}{\text{Therefore the answer is } \fbox{ 10000 - 9801 = 199}}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_} \\ \color{#3D99F6}{\text{Instead you can also use Arithmetico-Geometric progression.}} \\ \text{N is in AGP, where a = 1, d = 1, r = }\dfrac{1}{100} \\ \text{As AGP tends to infinity} \\ \implies \color{#3D99F6}{N = \dfrac{a}{1-r} + \dfrac{dr}{(1-r)^2}} \\ \implies \color{#3D99F6}{N = \dfrac{1}{1-\dfrac{1}{100}} + \dfrac{1\times\dfrac{1}{100}}{(1-\dfrac{1}{100})^2}} \\ \implies N = \dfrac{1}{\dfrac{100 - 1}{100}} + \dfrac{\dfrac{1}{100}}{(\dfrac{100-1}{100})^2} \\ \implies N = \dfrac{100}{99} + \dfrac{100}{99 \times 99} \\ \implies N = \dfrac{100 \times 99 + 100}{99 \times 99} \\ \implies N = \dfrac{10000}{9801} \\ \color{#D61F06}{\text{Therefore, a - b = } \fbox{ 199 }}$

$1000N=10.20304050......- - - - 1$
$10N=1020.30405060......- - - - 2$ [Subtract 2nd from the 1st equation]
$\text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}$
$990N=1010.\overline{10}\Rightarrow 990N=1010\dfrac{10}{99}$
Therefore $N=\dfrac{10000}{9801}.$ $a=10000$ and $b=9801.$ So $a-b=10000-9801=\boxed{199}.$

Method 1 - General Method

From Binomial - Theorem , $(1-x)^{-1} = 1+x+x^2+x^2 + \cdots = \sum_0^\infty x^i$

Differentiating, $\dfrac d{dx} \sum_0^\infty x^i = \sum_0^\infty ix^{i-1} = \dfrac d{dx} (1-x)^{-1} = (1-x)^{-2}$

Substituting $x=100$ , $N = \dfrac{100^2}{99^2}$

Hence $a-b = 100+99 = \color{#3D99F6}{\boxed{199}}$

Method 2 - Direct Binomial Theorem

From Binomial - Theorem , we get $(1-x)^{-n} = \sum_0^\infty \binom{n+r-1}rx^r$

Using transform $r+1\to R$ $(1-x)^{-n} = \sum_0^\infty \binom{n+R-2}{R-1}x^{R-1}$

We need $\binom{n+R-2}{R-1} = R$ . So $n=2$

Thus we get, the required sum is $N = (1-x)^{-2}~\text{at }~\left\{x=\dfrac1{100}\right\} = \dfrac{100^2}{99^2}$

Hence $a-b = 100+99 = \color{#3D99F6}{\boxed{199}}$

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Binomial Theorem

$(1+x)^n = \sum_0^n \binom n r x^r\\(1+x)^{p/q} = 1 + \dfrac{p}{1!}\left(\dfrac x q\right)^1+\dfrac{p(p-q)}{2!}\left(\dfrac x q\right)^2 + \dfrac{p(p-q)(p-2q)}{3!}\left(\dfrac x q\right)^3+\cdots\\ (1-x)^{-p/q} = 1 + \dfrac{p}{1!}\left(\dfrac x q\right)^1+\dfrac{p(p+q)}{2!}\left(\dfrac x q\right)^2 + \dfrac{p(p+q)(p+2q)}{3!}\left(\dfrac x q\right)^3+\cdots\to~\text{Remember only this}\\\begin{aligned} (1-x)^{-n} &= 1 + \dfrac{n}{1!}\left(\dfrac x 1\right)^1+\dfrac{n(n+1)}{2!}\left(\dfrac x 1\right)^2 + \dfrac{n(n+1)(n+2)}{3!}\left(\dfrac x 1\right)^3+\cdots\\&= \sum_0^\infty \binom{n+r-1}r x^r\end{aligned}$

Here you go, some links:

• Binomial Theorem
• Negative Binomial Theorem
• Fractional Binomial Theorem

$N = 1 + 0.02 + 0.0003 + 0.000004 + ... = 1 + 2(0.1)^2 + 3(0.1)^4 + 4(0.1)^6 + ...$

Let $x=0.1$ , then:

$N = 1 + 2(x)^2 + 3(x)^4 + 4(x)^6 + ... = 1 + 3x^2 - x^2 + 5x^4 - 2x^4 + 7x^6 - 3x^6 + ... = (1+3x^2+5x^4+7x^6+..) - (x^2 + 2x^4 + 3x^6 + ...)$

$= [\frac{d}{dx} (x + x^3 + x^5 + ..)] - [(x^2 + x^4 + x^6 + ..) + (x^4 + x^6 + ...) + (x^6 + x^8 + ..) + ...]$

Using the formula for infinite GP, we have:

$= \frac{d}{dx} (\frac{x}{1-x^2}) - ( \frac{x^2}{1-x^2} + \frac{x^4}{1-x^2} + \frac{x^6}{1-x^2} + ..)$

$= \frac{(1-x^2) \cdot 1 - (-2x) \cdot x}{(1-x^2)^2} - \frac{x^2 + x^4 + x^6 + x^8 + ...}{1-x^2}$

Once again, applying the infinite GP formula,

$= \frac{1+x^2}{(1-x^2)^2} - \frac{x^2}{(1-x^2)^2}$

$= \frac{1}{(1-x^2)^2}$

Now, we can substitute $x=0.1$ :

$= \frac{1}{(1-0.1^2)^2} = \frac{10000}{9801} \implies 10000-9801 = \boxed{199}$

This assumes that the progression of consecutive 2-digit numbers $02,03,04,05,....$ ends at $...95,96,97$ . That is, there is no $98$ , as the sequence goes $...95,96,97,99,00,01$ before it all repeats. That's the curious thing about this sequence, all of the 2 digit numbers from $00$ to $99$ are present, with the sole exception of $98$ .

In order to get the same thing but with $98$ included, we'd have to have the horrifically inelegant fraction of

$1030610152128364555667892062137547292123355790328548210397001336701367$
$3104888287012560045923988378839924600561270288849107337016734017040108$
$25529037956341292735538220692796756463729221611070402010000$

divided by

$1010101010101010101010101010101010101010101010101010101010101010101010$
$1010101010101010101010101010101010101010101010101010101010101010101010$
$10101010101010101010101010101010101010101010101010101010101$

What a difference that $98$ makes!