0

I have to show that $\frac{1}{9801}= 0.\overline{000102030405060708\dots9799}$. Here bar denotes Period.

My Attempt: I have shown $\frac{1}{9801}= {0.000102030405060708........9799.........}$ using the power series expansion of $\frac{1}{(1-x)^2}$. But I can not find the period of the decimal expansion of $\frac{1}{9801}$.

Can anyone help me with this ?

Thomas Andrews
  • 177,126
anonymous
  • 2,331
  • 14
  • 29

1 Answers1

0

If $n$ is an integer $\geqslant3,$ and $S$ is the sequence $0,1,2,\ldots,n-3,n-1,$ then the base $n$ expansion of the fraction $1/(n-1)^2$ is $(.\overline{S})_n,$ because: \begin{align*} (n-1)^2(.\overline{S})_n &= (n-1)^2\left(\frac1{n^2}+\frac2{n^3}+\cdots+\frac{n-3}{n^{n-2}}+ \frac{n-1}{n^{n-1}}\right) \left(1+\frac1{n^{n-1}}+\frac1{n^{2n-2}}+\cdots\right) \\ &= (n-1)\left(\frac1n+\frac1{n^2}+\cdots+\frac1{n^{n-3}}+ \frac2{n^{n-2}}-\frac{n-1}{n^{n-1}}\right) \left(1-\frac1{n^{n-1}}\right)^{-1} \\ &= (n-1)\left(\frac1n+\frac1{n^2}+\cdots+\frac1{n^{n-3}}+ \frac{n+1}{n^{n-1}}\right) \left(\frac{n^{n-1}}{n^{n-1}-1}\right) \\ &= \left(1-\frac1{n^{n-3}}+\frac{n^2-1}{n^{n-1}}\right) \left(\frac{n^{n-1}}{n^{n-1}-1}\right) \\ &= \frac{n^{n-1}-n^2+n^2-1}{n^{n-1}-1} \\ &= 1. \end{align*}

Taking $n=100,$ we get: $$ 1/9801 = (.\overline{00,01,02,\ldots,97,99})_{100} = .\overline{00010203\ldots95969799}, $$ so $1/9801$ has period 198 in base $10.$

[But see my comment on the meaning of the word "period" in the question.]

Calum Gilhooley
  • 12,174
  • With hindsight, it would have been tidier to write: \begin{align} (n-1)^2\left(\frac1{n^2}+\frac2{n^3}+\cdots+ \frac{n-3}{n^{n-2}}+\frac{n-1}{n^{n-1}}\right) &= (n-1)\left(\frac1n+\frac1{n^2}+\cdots+\frac1{n^{n-3}}+ \frac2{n^{n-2}}-\frac{n-1}{n^{n-1}}\right) \ &= (n-1)\left(\frac1n+\frac1{n^2}+\cdots+\frac1{n^{n-3}}+ \frac{n+1}{n^{n-1}}\right) \ &= 1-\frac1{n^{n-3}}+\frac{n^2-1}{n^{n-1}} = 1-\frac1{n^{n-1}}. \end{align} In particular, taking $n=100$: $$ 9801\times.00010203\ldots95969799 = .\underbrace{99999999\ldots99999999}_{198 \text{ digits}}. $$ – Calum Gilhooley Dec 29 '23 at 18:02
  • Alternatively, but really quite similarly: if $n\geqslant3,$ then \begin{align} n^{n-1}-1&=(n-1)\sum_{j=0}^{n-2}n^j\&= (n-1)\bigg[(n-1)+\sum_{j=1}^{n-2}(n^j-1)\bigg]\&= (n-1)^2\bigg[1+\sum_{j=1}^{n-2}\sum_{k=0}^{j-1}n^k\bigg]\&= (n-1)^2\bigg[1+\sum_{k=0}^{n-3}\sum_{j=k+1}^{n-2}n^k\bigg]\&= (n-1)^2\bigg[1+\sum_{k=0}^{n-3}(n-k-2)n^k\bigg]\&= (n-1)^2\bigg[(n-1)+\sum_{k=1}^{n-3}(n-k-2)n^k\bigg]. \end{align} For example, $999999999=81\times12345679.$ The point is that the expansion of $1/(n-1)^2$ in base $n$ follows from the above identity which only involves integers. – Calum Gilhooley Dec 30 '23 at 15:30