標題:
10points 1 math question
http://postimg.org/image/zawvu8nar/ http://postimg.org/image/rw7k1v1f7/ only part b) thanks a lot.
最佳解答:
bi) As the sum of n terms is (1－1/2^n), and,the sum of 2n terms is [1－1/2^(2n)]. So,{[1－1/2^(2n)]－(1－1/2^n)}/[1－1/2^(2n)] = 1/33==> [2^(2n)－1－2^(2n)＋2^n]/[2^(2n)－1]＝1/33==> 33*2^n－33＝2^(2n)－1==> 2^(2n)－33*2^n＋32＝0 bii) 2^(2n)－33*2^n＋32＝0==> (2^n－1)(2^n－32)＝0==> 2^n＝1 or 2^n＝32==> n＝0 (rej) or n＝5∴ n＝5 2015-04-05 13:25:00 補充： Sorry, (bi) part 計錯啲sum, 但冇影響個答案. Anyway, 再計過： bi) As the sum of n terms is 2(1－1/2^n), and, the sum to 2n terms is 2[1－1/2^(2n)]. So, {2[1－1/2^(2n)]－2(1－1/2^n)}/{2[1－1/2^(2n)]} = 1/33 ==> [2^(2n)－1－2^(2n)＋2^n]/[2^(2n)－1]＝1/33 ==> 33*2^n－33＝2^(2n)－1 ==> 2^(2n)－33*2^n＋32＝0
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10points 1 math question
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發問:http://postimg.org/image/zawvu8nar/ http://postimg.org/image/rw7k1v1f7/ only part b) thanks a lot.
最佳解答:
bi) As the sum of n terms is (1－1/2^n), and,the sum of 2n terms is [1－1/2^(2n)]. So,{[1－1/2^(2n)]－(1－1/2^n)}/[1－1/2^(2n)] = 1/33==> [2^(2n)－1－2^(2n)＋2^n]/[2^(2n)－1]＝1/33==> 33*2^n－33＝2^(2n)－1==> 2^(2n)－33*2^n＋32＝0 bii) 2^(2n)－33*2^n＋32＝0==> (2^n－1)(2^n－32)＝0==> 2^n＝1 or 2^n＝32==> n＝0 (rej) or n＝5∴ n＝5 2015-04-05 13:25:00 補充： Sorry, (bi) part 計錯啲sum, 但冇影響個答案. Anyway, 再計過： bi) As the sum of n terms is 2(1－1/2^n), and, the sum to 2n terms is 2[1－1/2^(2n)]. So, {2[1－1/2^(2n)]－2(1－1/2^n)}/{2[1－1/2^(2n)]} = 1/33 ==> [2^(2n)－1－2^(2n)＋2^n]/[2^(2n)－1]＝1/33 ==> 33*2^n－33＝2^(2n)－1 ==> 2^(2n)－33*2^n＋32＝0
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