Some useful formulae

For any number r,

1 + r + r² + ... + rⁿ = SUM(k=0...n) r^k = (1 - rⁿ⁺¹) /(1 - r), and
r + r² + r³ + ... + rⁿ = SUM(k=1...n) r^k = (r - rⁿ⁺¹)/(1 - r).

Thus, if |r| < 1 then taking the limit n -> infinity,

1 + r + r² + r³ + . . . = SUM(k=0... infinity) r^k = 1/(1-r), and
r + r² + r³ + . . . = SUM(k=1...infinity) r^k = r/(1-r).

Combining the above formulae, we obtain,

r^m + r^m+1 + ... + r^m+k = r^m(1 + r + r² + . . . + r^k) = (r^m - r^m+k+1) / (1 - r ), for any r, and
r^m + r^m+1 + r^m+2 + . . . = SUM(k=m... infinity) r^k = r^m(1 + r + r² + . . . ) = r^m /(1 - r) for |r| < 1.

Let x be any number in [0,1], and [x]_a denote the expansion of x (or the representation of x) in base a (a can be any positive integer).

Decimal expansion (a=10):

[x]₁₀=.d₁d₂d₃. . . , where d_i= 0, 1, 2, ... or 9.
Then x = d₁/10 + d₂/10² + d₃/10³. . .

Binary expansion (a = 2):

[x]₂=.b₁b₂b₃. . . , where b_i= 0 or 1.
Then x = b₁/2 + b₂/2² + b₃/2³. . .

Ternary expansion (a = 3):

[x]₃=.c₁c₂c₃. . . , where c_i= 0, 1 or 2.
Then x = c₁/3 + c₂/3² + c₃/3³. . .

For numbers greater than 1, there would be digits to the left of the point for the positive powers of a.
Note that some numbers may have two expansions. For example, [1/10]₁₀ = .1 and .099(99). Also, [3/4]₂=.11 and also .1011(11). Another, [7/9]₃=.021 and also .02022(22) (this can be seen by adding up the infinite series using the geometric formula above).
(Here, (ab) denotes the repeating pattern abababab..... )

Here's the algorithm for computing the ternary expansion of a number x between 0 and 1 (there's a similar algorithm for other bases and if the number is greater than 1). Here we want to determine c₁, c₂, c₃, . . . where each c_i is either 0, 1, or 2 such that
x = c₁/3 + c₂/3² + c₃/3³ + ... .

Determine c₁;

• if 0 < x < 1/3, then c₁ = 0
• if 1/3 < = x < 2/3, then c₁ = 1
• if 2/3 < = x < 1, then c₁ = 2
• let x₁ = x - c₁/3 (note that x₁ < 1/3 )

Determine c₂;

• if 0 < x₁ < 1/3², then c₂ = 0
• if 1/3² < = x < 2/3², then c₂ = 1
• if 2/3² < = x < 1/3, then c₂ = 2
• let x₂ = x₁ - c₂/3² (note that x₂ < 1/3²)

Determine c₃;

• if 0 < x₂ < 1/3³, then c₃ = 0
• if 1/3³ < = x₂ < 2/3³, then c₃ = 1
• if 2/3³ < = x₂ < 1/3², then c₃ = 2
• let x₃ = x₂ - c₃/3³ (note that x₃ < 1/3³)

etc.