看板 Math 關於我們 聯絡資訊
※ 引述《izzylf (風)》之銘言: : 各位版友好 : 這裡有幾題數列級數和指對數問題請教 : 因為符號不好在bbs顯現,所以用printscreen的 : 圖貼在自己的相簿 : 謝謝幫我回答的版友 : http://www.wretch.cc/album/show.php?i=izzylf0612&b=3&f=1619378913&p=1 1. S = 1 + (1+2) + (1+2+2^2) + ... + [1+2+2^2+...+2^(n-1)] n-1 n-1 = Σ[2^(k+1)-1] = Σ2^(k+1) - n = 2^(n+1) - 2 - n k=0 k=0 2. S_n = 1/1 + 1/(1+2) + ... + 1/(1+2+...+n) n n = Σ2/[(k+1)k] = 2Σ[1/k - 1/(k+1)] = 2 - 2/(n+1) → 2 as n→∞ k=1 k=1 |S-S_n|<0.01 => 2/(n+1)<0.01 => n > 2/0.01 - 1 = 199 n最小值=200 3. a = log_60(3), b = log_60(5) [1-a-b]/[2-2b] = [1-log_60(3)-log_60(5)]/[2-2log_60(5)] = log_60[60/(3x5)]/log_60[3600/25] = log_60(4)/log_60(144) = log_12(4)/log_12(144) = log_12(2) Thus, 12^[(1-a-b)/(2-2b)] = 2 4. a^2 - 1 = (1/4)[47^(2/2010)+47^(-2/2010)]+1/2-1 = {(1/2)[47^(1/2010)-47^(-1/2010)]}^2 Then, Ans = [47^(-1/2010)]^2010 = 1/47 -- ~by Jackary P.~ -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 112.78.69.240
izzylf :感謝感謝!! 04/23 21:53