Hãy Tính Tổng:
C=(1-1/3)*(1-1/6)*(1-1/9)*(1-1/15)*...................*(1-1/780)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
(1-1/3) x (1-1/6) x (1-1/10) x (1-1/15)x ...x (1-1/780)
=2/3x 5/6 x 9/10 x...x 779/780
=4/6 x 10/12 x 18/20 x ...x 1558/1560
=4x10 x 18 x...x 1558/6x 12 x 20 x ...x 1560
= 41/39x 3
= 41/11
\(B=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)......\left(1-\frac{1}{780}\right)\)
\(B=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}......\frac{779}{780}\)\(=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}.....\frac{1558}{1560}\)
\(B=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.....\frac{38.41}{39.40}\)
\(B=\frac{\left(1.2.3.....38\right)\left(4.5.6.....41\right)}{\left(2.3.4.....39\right)\left(3.4.5.....40\right)}=\frac{1.41}{39.3}=\frac{41}{117}\)
\(B=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)......\left(1-\frac{1}{780}\right)\)
\(\Rightarrow B=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}......\frac{779}{780}=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}.....\frac{1558}{1560}\)
\(\Rightarrow B=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.....\frac{38.41}{39.40}\)
\(\Rightarrow B=\frac{\left(1.2.3.....38\right)\left(4.5.6.....41\right)}{\left(2.3.4.....39\right)\left(3.4.5.....40\right)}=\frac{1.41}{39.3}=\frac{41}{117}\)
mk chỉnh lại đề
\(A=\frac{8}{9}.\frac{15}{16}.\frac{24}{25}....\frac{99}{100}\)
\(=\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}.\frac{5^2-1}{5^2}....\frac{10^2-1}{10^2}\)
\(=\frac{2.4}{3^2}.\frac{3.5}{4^2}.\frac{4.6}{5^2}.....\frac{9.11}{10^2}\)
\(=\frac{2.3.4...9}{3.4.5...10}.\frac{4.5.6...11}{3.4.5...10}\)
\(=\frac{2}{10}.\frac{11}{3}=\frac{11}{15}\)
\(D=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)......\left(1-\frac{1}{780}\right)\)
\(\Rightarrow D=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}......\frac{779}{780}=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}.....\frac{1558}{1560}\)
\(\Rightarrow D=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.....\frac{38.41}{39.40}\)
\(\Rightarrow D=\frac{\left(1.2.3.....38\right)\left(4.5.6.....41\right)}{\left(2.3.4.....39\right)\left(3.4.5.....40\right)}=\frac{1.41}{39.3}=\frac{41}{117}\)
3:
\(A=10^{15}+5=1000...05\)(Có 15 chữ số 0)
Tổng các chữ số trong số A là:
1+0+0+...+0+5=6
=>A chia hết cho 3
=>Số dư khi A chia cho 3 là 0
Vì tổng các chữ số trong A là 6 không chia hết cho 9
nên số dư của A khi chia cho 9 là 6
5:
Số số hạng trong dãy từ 4 đến 160 là: \(\dfrac{160-4}{4}+1=\dfrac{156}{4}+1=40\left(số\right)\)
Tổng các số trong dãy từ 4 đến 160 là:
\(\left(160+4\right)\cdot\dfrac{40}{2}=164\cdot20=3280\)
=>C=3280+1=3281
\(B=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)......\left(1-\frac{1}{780}\right)\)
\(B=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}......\frac{779}{780}\)\(=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}.....\frac{1558}{1560}\)
\(B=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.....\frac{38.41}{39.40}\)
\(B=\frac{\left(1.2.3.....38\right)\left(4.5.6.....41\right)}{\left(2.3.4.....39\right)\left(3.4.5.....40\right)}=\frac{1.41}{39.3}=\frac{41}{117}\)
a)\(\frac{8}{9}.\frac{15}{16}.\frac{24}{25}.....\frac{2499}{2500}\) \(=\frac{2.4}{3^2}.\frac{3.5}{4^2}.\frac{4.6}{5^2}.....\frac{49.51}{50^2}\)\(=\frac{\left(2.3.4.....49\right)\left(4.5.6....51\right)}{\left(3.4.5.....50\right)\left(3.4.5.....50\right)}=\frac{2.51}{50.3}=\frac{1.17}{25}=\frac{17}{25}\)
b)\(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)......\left(1-\frac{1}{780}\right)\)
\(=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}......\frac{779}{780}\)\(=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}.....\frac{1558}{1560}\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.....\frac{38.41}{39.40}\)
\(=\frac{\left(1.2.3.....38\right)\left(4.5.6.....41\right)}{\left(2.3.4.....39\right)\left(3.4.5.....40\right)}=\frac{1.41}{39.3}=\frac{41}{117}\)
\(\left(1-\frac{1}{3}\right).\left(1-\frac{1}{6}\right).\left(1-\frac{1}{10}\right).\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{780}\right)\)
\(=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}...\frac{779}{780}\)
\(=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}.\frac{28}{30}...\frac{1558}{1560}\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{4.7}{5.6}...\frac{38.41}{39.40}\)
\(=\frac{1.2.3.4...38}{2.34.5...39}.\frac{4.5.6.7...41}{3.4.5.6...40}\)
\(=\frac{1}{39}.\frac{41}{3}=\frac{41}{297}\)
Trương Quang Minh