Reciprocal of 98
Jump to navigation
Jump to search
Theorem
The decimal expansion of the reciprocal of $98$ starts with the powers of $2$:
- $\dfrac 1 {98} = 0 \cdotp 0 \dot 1020 \, 40816 \, 32653 \, 06122 \, 44897 \, 95918 \, 36734 \, 69387 \, 75 \dot 5$
This sequence is A021102 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).
Proof
Performing the calculation using long division:
0.01020408163265306122448979591836734693877551...
--------------------------------------------------
98)1.00000000000000000000000000000000000000000000000
98 196 196 490 294 686
-- --- --- --- --- ---
200 640 440 900 460 540
196 588 392 882 392 490
--- --- --- --- --- ---
400 520 480 180 680 500
392 490 392 98 588 490
--- --- --- --- --- ---
800 300 880 820 920 100
784 294 784 784 882 98
--- --- --- --- --- ---
160 600 960 360 380 ...
98 588 882 294 294
--- --- --- --- ---
620 120 780 660 860
588 98 686 588 784
--- --- --- --- ---
320 220 940 720 760
294 196 882 686 686
--- --- --- --- ---
260 240 580 340 740
196 196 490 294 686
We have that:
- $0 \cdotp 01 + 0 \cdotp 0002 + 0 \cdotp 000004 + 0 \cdotp 00000008 + \cdots$
is nothing else but:
- $\ds \dfrac 1 2 \paren {\dfrac 1 {50} + \paren {\dfrac 1 {50} }^2 + \paren {\dfrac 1 {50} }^3 + \paren {\dfrac 1 {50} }^4 + \cdots} = \dfrac 1 2 \sum_{k \mathop \ge 1} \paren {\dfrac 1 {50} }^k$
Hence:
| \(\ds \dfrac 1 2 \sum_{k \mathop \ge 1} \paren {\dfrac 1 {50} }^k\) | \(=\) | \(\ds \dfrac 1 2 \dfrac {1 / 50} {1 - 1 / 50}\) | Sum of Infinite Geometric Sequence: Corollary 1 | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 1 2 \dfrac 1 {50 - 1}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 1 2 \dfrac 1 {49}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 1 {98}\) |
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $98$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $98$