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Please help!!! What is the units digit of 3^1+3^3+3^5 ... 3^2009?
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\[3^1 +3^3+ 3^5+3^7 \cdots 3^{2009}\]
Notice that this is \[\sum_{n = 1}^{2009}3^{2n-1}\] And by plugging in a few you can notice a pattern: 3 + 27 + 243 +2187 + 19683 the unit of any partial sum alternates from 3 to 0 notice that when n = odd number. the unit of that term is 3 when n = even number, the unit of that term is 7, but when added to the previews gives 0 [from 10] so is 2009 a term where n is odd or even? find out: 2n - 1 = 2009 n = 1005 = odd so the unit of that "partial sum", which is the entire sum is 3
previous* lol let me know if you have any follow-up questions about by work
oh!!! Thanks!
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glad i could help :)
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