a-n-1-n-n-9-1-n-0-a-n-b-n-a-n-1-9-a-n-n-1-n-0-b-n-

Question Number 2329 by 123456 last updated on 16/Nov/15

a_{n} = {(- 1)}^{n} (⌊ \frac{n}{9} ⌋ + 1)

\sum_{n ⩾ 0} a_{n} = ?

b_{n} = \frac{(a_{n} - 1) (9 - a_{n})}{n + 1}

\sum_{n ⩾ 0} b_{n} = ?

Commented by prakash jain last updated on 16/Nov/15

a_{0} = 1, a_{1} = - 1, a_{2} = 1 . ., a_{8} = 1, \underset{n = 0}{\sum^{8}} a_{n} = 1

a_{9} = 2, a_{10} = - 2, a_{11} = 2 . ., a_{17} = 2, \underset{n = 9}{\sum^{n = 17}} a_{n} = 2

⌊ \frac{n}{9} ⌋ = k

n - k = m

\underset{i = 0}{\sum^{n}} a_{i} = \frac{k (k + 1)}{2} + {(- 1)}^{m} \frac{1 + k}{2} + \frac{1 + k}{2}

Commented by Rasheed Soomro last updated on 16/Nov/15

⌊ \frac{n}{9} ⌋ b r a c k e t f u n c t i o n ?

Commented by 123456 last updated on 16/Nov/15

greatest lower integer

⌊ 0, 3 ⌋ = 0

⌊ 0, 7 ⌋ = 0

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