Octal number equations -

so i'm reading book on how binary bits converted octal numbers. when trying explain concept, give equation

n= s(...((d8)2^8+(d7)2^7+(d6)2^6)+((d5)2^5+(d4)2^4+(d3)2^3)+((d2)2^2+(d1)2^1+d0))

or

n= s(...((d8)2^2 +(d7)2+(d6))2^6 + ((d5)2^2 +(d4)2^1 + (d3))2^3 + ((d2)2^2+(d1)2^1+d0))

d represents digit found within bit, e.g. if least significant bit 1, (d0) 1.

i understand of this, elaborate further saying parenthesized expressions ((d8)2^2 +(d7)2+(d6)) coefficients of base 8 digits, n=s((d2)8^2+(d1)*8+(d0)).

can explain mean parenthesized expressions being coefficients of base8 digits?

the digits d_i binary digits of number. can compute number binary digits this:

    n = ∑ _i 2ⁱ d_i = 2⁰ d₀ + 2¹ d₁ + 2² d₂ + ⋯

(this in fact defines “binary”, if add condition digits integers , 0 ≤ d_i < 2 i.)

suppose name octal digits of number o_j. can compute number octal digits this:

    n = ∑ _j 8^j o_j = 8⁰ o₀ + 8¹ o₁ + 8² o₂ + ⋯

(this defines “octal”, if add condition digits integers , 0 ≤ o_j < 8 j.)

now let's @ binary equation. first step trickiest. change way subscript used each term of summation uses 3 binary digits:

    n = ∑ _j 2^{3 j + 0} d_{3 j + 0} + 2^{3 j + 1} d_{3 j + 1} + 2^{3 j + 2} d_{3 j + 2}

convince that equation computes same n first equation gave.

i assume know x^{a + b} = x^a x^b. can separate 2^{3 j + b} coefficients this:

    n = ∑ _j (2^{3 j} 2⁰) d_{3 j + 0} + (2^{3 j} 2¹) d_{3 j + 1} + (2^{3 j} 2²) d_{3 j + 2}

then can factor out 2^{3 j} term this:

    n = ∑ _j 2^{3 j} (2⁰ d_{3 j + 0} + 2¹ d_{3 j + 1} + 2² d_{3 j + 2})

i assume know x^{a b} = (x^a)^b. can split 2^{3 j} term this:

    n = ∑ _j (2³)^j (2⁰ d_{3 j + 0} + 2¹ d_{3 j + 1} + 2² d_{3 j + 2})

and can simplify 2³ 8:

    n = ∑ _j 8^j (2⁰ d_{3 j + 0} + 2¹ d_{3 j + 1} + 2² d_{3 j + 2})

compare formula computing number octal digits, repeat here:

    n = ∑ _j 8^j o_j

so can conclude this:

    o_j = 2⁰ d_{3 j + 0} + 2¹ d_{3 j + 1} + 2² d_{3 j + 2}

for example, let's take j = 2:

    o₂ = 2⁰ d_{3×2 + 0} + 2¹ d_{3×2 + 1} + 2² d_{3×2 + 2} = 2⁰ d₆ + 2¹ d₇ + 2² d₈