# It's Not Magic

$\dfrac{19}{81}\approx 0.\red{2}\blue{3}\red{4}\blue{5}\red{6}\blue{7}\red{9}\red{0}\blue{1}...$ $\dfrac{199}{9801} \approx 0.\red{02}\blue{03}\red{04}\blue{05}\red{06}\blue{07}\red{08}\blue{09}\red{10}\blue{11}...$ $How\space to\space generate\space such\space rationals?$ $let \space there \space be \space a \space sequence \space a_n$ $a_1=\alpha\beta$ $\forall n\in\mathbb{N},a_{n+1}=(\alpha+nd)\beta r^n$ $let \space S_n = \sum_{k=1}^{n}a_k$ $=\sum_{k=1}^{n} (\alpha+(k-1)d)\beta r^{k-1}$ $=\sum_{k=1}^{n} (\alpha\beta r^{k-1}+(k-1)d\beta r^{k-1})$ $=\alpha\beta\sum_{k=1}^{n}r^{k-1}+\beta d\sum_{k=1}^{n}(k-1)r^{k-1}$ $=\alpha\beta\dfrac{r^{n}-1}{r-1}-\beta d\sum_{k=1}^{n}r^{k-1}+\beta d\blue{\sum_{k=1}^{n}kr^{k-1}}$ $=\beta\dfrac{r^{n}-1}{r-1}(\alpha-d)+\beta d\blue{S}$ $\blue{S} =\sum_{k=1}^{n}kr^{k-1}$ $S\times r=\sum_{k=1}^{n}kr^{k}=nr^n+\sum_{k=2}^{n}(k-1)r^{k-1}$ $\Rightarrow S(1-r)=-nr^n+\sum_{k=1}^{n-1}r^k$ $=\dfrac{r^n-1}{r-1}-nr^n$ $\Rightarrow S=\dfrac{r^n-1}{(r-1)^2}-\dfrac{nr^n}{r-1}$ $Putting \space this\space we\space get$ $S_n=\dfrac{\beta(1-r^n)(\alpha(1-r)+dr)}{(1-r)^2}-\dfrac{\beta dnr^n}{1-r}$ $\Rightarrow \sum_{k=1}^{\infty}a_k=\dfrac{\beta(\alpha(1-r)+dr)}{(1-r)^2}$ $Put\space \alpha=1=\beta=d, r=10^{-n},n\in\mathbb{N}$ $\Rightarrow \sum_{k=1}^{\infty}a_k=\dfrac{100^n}{(10^n-1)^2}$ $Now\space consider B_n=\dfrac{100^n}{(10^n-1)^2}-1$ $\Rightarrow \boxed{B_n=\dfrac{2\times 10^n - 1}{(10^n - 1)^2}}$ $Which\space is \space the \space Formula \space we \space wanted!$

Note by Zakir Husain
3 months, 3 weeks ago

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