Given-that-N-625-1-9-5-2-1-9-8-2-1-9-11-2-1-9-125-2-Find-the-sum-of-the-digits-of-N-

Question Number 114570 by ZiYangLee last updated on 19/Sep/20

Given that

N = 625 (1 - \frac{9}{5^{2}}) (1 - \frac{9}{8^{2}}) (1 - \frac{9}{11^{2}}) \dots (1 - \frac{9}{125^{2}})

Find the sum of the digits of N .

Answered by floor(10²Eta[1]) last updated on 19/Sep/20

N = 625 (\frac{2.8}{5^{2}}) (\frac{5.11}{8^{2}}) (\frac{8.14}{11^{2}}) (\frac{11.17}{14^{2}}) \dots (\frac{122.128}{125^{2}})

N = 625 (\frac{2}{5}) (\frac{128}{125}) = 2.128 = 256

2 + 5 + 6 = 13

Answered by Olaf last updated on 19/Sep/20

N = 625 \underset{n = 1}{\prod^{41}} (1 - \frac{9}{{(3 n + 2)}^{2}})

N = 625 \underset{n = 1}{\prod^{41}} (\frac{{(3 n + 2)}^{2} - 9}{{(3 n + 2)}^{2}})

N = 625 \underset{n = 1}{\prod^{41}} (\frac{9 n^{2} + 12 n - 5}{{(3 n + 2)}^{2}})

9 n^{2} + 12 n - 5 = 0

Δ^{'} = 36 - 9 (- 5) = 36 + 45 = 81

1 st root : \frac{- 6 - 9}{9} = - \frac{5}{3}

2 nd root : \frac{- 6 + 9}{9} = \frac{1}{3}

N = 625 \underset{n = 1}{\prod^{41}} \frac{9 (n + \frac{5}{3}) (n - \frac{1}{3})}{{(3 n + 2)}^{2}}

N = 625 \underset{n = 1}{\prod^{41}} \frac{(3 n + 5) (3 n - 1)}{{(3 n + 2)}^{2}}

N = 625 \underset{n = 1}{\prod^{41}} \frac{(3 (n + 1) + 2) (3 (n - 1) + 2)}{{(3 n + 2)}^{2}}

u_{n} = 3 n + 2, n ⩾ 0

N = 625 \frac{\underset{n = 1}{\prod^{41}} u_{n + 1} \times \underset{n = 1}{\prod^{41}} u_{n - 1}}{{(\underset{n = 1}{\prod^{41}} u_{n})}^{2}}

N = 625 \frac{\underset{n = 2}{\prod^{42}} u_{n} \times \underset{n = 0}{\prod^{40}} u_{n}}{{(\underset{n = 1}{\prod^{41}} u_{n})}^{2}}

N = 625 \times \frac{u_{42}}{u_{1}} \times \frac{u_{0}}{u_{41}}

u_{0} = 3 (0) + 2 = 2

u_{1} = 3 (1) + 2 = 5

u_{41} = 3 (41) + 2 = 125

u_{42} = 3 (42) + 2 = 128

N = 625 \times \frac{128}{5} \times \frac{2}{125}

N = 256

sum of the digits of N is :

2 + 5 + 6 = 13

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