Reciprocal of 49 shows Powers of 2 in Decimal Expansion

Theorem
The decimal expansion of the reciprocal of $49$ contains the powers of $2$:
 * $\dfrac 1 {49} = 0 \cdotp \dot 02040 \, 81632 \, 65306 \, 12244 \, 89795 \, 91836 \, 73469 \, 38775 \, 5 \dot 1$

Proof
From Reciprocal of $49$:

Adding up powers of $2$, shifted appropriately to the right: 02 04    08      16        32          64           128             256               512                1024                  2048                    4096                      8192                       16384                         32768                           65536                            131072                              262144                                524288                                 1048576                                   2097152                                     4194304                                       8388608                                        16777216                                          33554432                                            67108864                                             134217728                                               268435456                                                 536870912                                                  1073741824                                                    2147483648                                                      4294967296                                                        8589934592                                                         ...........

020408163265306122448979591836734693877551020408163265305947144192

1 1 1111 11 11212122222112222222322222111

As can be seen, the decimal expansion of $\dfrac 1 {49}$ matches the sum of the shifted powers of $2$ (to the limits of calculation) and the pattern is apparent.

This is what is to be expected, because:
 * $0 \cdotp 02 + 0 \cdotp 0004 + 0 \cdotp 000008 + 0 \cdotp 00000016 + \cdots$

is nothing else but:
 * $\dfrac 1 {50} + \paren {\dfrac 1 {50} }^2 + \paren {\dfrac 1 {50} }^3 + \paren {\dfrac 1 {50} }^4 + \cdots = \ds \sum_{k \mathop \ge 1} \paren {\dfrac 1 {50} }^k$

Hence: