Expert Answer to "Why sum of digits in even base have..."

ilybabeilyl85

Answered question

2022-03-08

Why sum of digits in even base have such property?
Function

D (a, b)

define as sum of digit of a in base b. Example

D (5, 2) = 2

.
Let

f (m, n) = \sum_{k = 1}^{m} {(- 1)}^{D (k, n)}

Pentyrch9ci

Beginner2022-03-09Added 1 answers

Step 1
The point is that if

e l l, n

are even then

{(- 1)}^{D (e l l, n)} + {(- 1)}^{D (e l l + 1, n)} = 0

This is because

e l l

and

e l l + 1

differ only in the last digit in base n.
That means

\sum_{k = 0}^{e l l + 1} {(- 1)}^{D (k, n)} = 0

for

e l l

even, because it can be split into

e l l + 1

pairs which each sum to 0.
Now if

m = e l l + 1

is odd then

f (m, n) = \sum_{k = 0}^{e l l + 1} {(- 1)}^{D (k, n)} - {(- 1)}^{D (0, n)} = 0 - 1 = - 1,

whereas if

m = e l l

is even then

f (m, n) = \sum_{k = 0}^{e l l + 1} {(- 1)}^{D (k, n)} - {(- 1)}^{D (0, n)} - {(- 1)}^{D (e l l + 1, n)} = - 1 - {(- 1)}^{D (e l l + 1, n)},

and since

{(- 1)}^{D (e l l + 1, n)} \in {- 1, 1}

, we get

f (m, n) \in {- 2, 0}

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