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Ta có :
\(S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)
\(2S=6+3+\frac{3}{2}+...+\frac{3}{2^8}\)
\(2S-S=\left(6+3+\frac{3}{2}+...+\frac{3}{2^8}\right)-\left(3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\right)\)
\(S=6-\frac{3}{2^9}\)
\(S=\frac{2^{10}.3-3}{2^9}\)
Vậy \(S=\frac{2^{10}.3-3}{2^9}\)
vận dụng 3S lên
xong tìm S nha bn ok
tại k có thời gian nên chỉ giúp thế thôi
Bài 1:
a)\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(=1-\frac{1}{2017}\)
\(=\frac{2016}{2017}\)
b)\(=1008\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(=1008\cdot\left(1-\frac{1}{2017}\right)\)
\(=1008\cdot\frac{2016}{2017}\)\(=\frac{290304}{31}\)Bài 2:
a)\(A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{19.21}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\)
\(=\frac{1}{3}-\frac{1}{21}\)
\(=\frac{2}{7}\)
b)\(B=\frac{5}{28}+\frac{5}{70}+...+\frac{5}{700}\)
\(=\frac{5}{4.7}+\frac{5}{7.10}+...+\frac{5}{25.28}\)
\(=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{28}\right)\)
\(=\frac{5}{3}\cdot\frac{6}{28}\)
\(=\frac{15}{14}\)
Bài 3:
a)Đặt \(A=-\frac{20}{11.13}-\frac{20}{13.15}-...-\frac{20}{53.55}\)
\(=-\left(\frac{20}{11.13}+\frac{20}{13.15}+...+\frac{20}{53.55}\right)\)
\(=-\left[10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)\right]\)
\(=-\left[10\left(\frac{1}{11}-\frac{1}{55}\right)\right]\)
\(=-\left[10\cdot\frac{4}{55}\right]\)
\(=-\frac{8}{11}\).Thay vào ta có: \(x-\frac{8}{11}=\frac{2}{9}\)
\(\Leftrightarrow x=\frac{94}{99}\)
b)\(\frac{2}{42}+\frac{2}{56}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(2\left(\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{18}\)
\(x+1=18\)
\(x=17\)
\(E=\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+...+\frac{1}{73\cdot75}\)
\(E=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)
\(\Rightarrow E=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)=\frac{1}{2}\cdot\frac{2}{75}=\frac{1}{75}\)
\(F=\frac{15}{90\cdot94}+\frac{15}{94\cdot98}+...+\frac{15}{146\cdot150}\)
\(F=\frac{15}{4}\cdot\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+...+\frac{1}{146}-\frac{1}{150}\right)\)
\(\Rightarrow F=\frac{15}{4}\cdot\left(\frac{1}{90}-\frac{1}{150}\right)=\frac{15}{4}\cdot\frac{1}{225}=\frac{1}{60}\)
\(G=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
\(G=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)
\(G=\frac{5}{4\cdot7}+\frac{5}{7\cdot10}+\frac{5}{10\cdot13}+...+\frac{5}{25\cdot28}\)
\(G=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(\Rightarrow G=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{28}\right)=\frac{5}{3}\cdot\frac{3}{14}=\frac{5}{14}\)
a) A=(100-1):1+1=100 số hạng
A=100:2=50 cặp
tính giá trị của từng cặp số = (1+100)+(2+99)+(3+98)+...+(50+51)=101
tính giá trị của biểu thức A: 50*101=5050
[ mình tính theo công thức đó ]
1) \(D=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+....+\frac{10}{1400}\)
\(D=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+.....+\frac{5}{700}\)
\(D=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+......+\frac{5}{25.28}\)
\(D=\frac{5}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+.....+\frac{3}{25.28}\right)\)
\(D=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+....+\frac{1}{25}-\frac{1}{28}\right)\)
\(D=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{28}\right)=\frac{5}{3}.\frac{6}{28}=\frac{5}{14}\)
\(E=\frac{1}{1+2}+\frac{1}{1+2+3}+.......+\frac{1}{1+2+3+....+24}\)
Ta có: \(1+2=\)\(\frac{2.\left(2+1\right)}{2}=3\);\(1+2+3=\frac{3.\left(3+1\right)}{2}=6\);\(1+2+3+...+24=\frac{24.\left(24+1\right)}{2}=300\)
\(E=\frac{1}{3}+\frac{1}{6}+....+\frac{1}{300}\)
=>\(\frac{1}{2}E=\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{600}=\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{24.25}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{24}-\frac{1}{25}=\frac{1}{2}-\frac{1}{25}=\frac{23}{50}\)
=>\(E=\frac{46}{50}\)
Vậy \(\frac{D}{E}=\frac{5}{14}:\frac{46}{50}=\frac{250}{644}=\frac{125}{322}\)
2) Theo t/c dãy tỉ số=nhau:
\(\frac{a+b}{a+c}=\frac{a-b}{a-c}=\frac{a+b-\left(a-b\right)}{a+c-\left(a-c\right)}=\frac{a+b-a+b}{a+c-a+c}=\frac{2b}{2c}=1\)
=>b=c
do đó \(A=\frac{10b^2+9bc+c^2}{2b^2+bc+2c^2}=\frac{10b^2+9b^2+b^2}{2b^2+b^2+2b^2}=\frac{\left(10+9+1\right).b^2}{\left(2+1+2\right).b^2}=4\)
Bài 1 mik học xong quên hết òi (mấy bài kia là hok biết luôn :V)
\(\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+....+\frac{10}{1400}\)
\(=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+.....+\frac{5}{25.28}\)
\(=\frac{5}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+....+\frac{3}{25.28}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{25}-\frac{1}{28}\right)\)
\(=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{28}\right)\)
\(=\frac{5}{3}.\frac{3}{14}=\frac{5}{14}\)
a)Đặt A=Tổng trên, ta có:
\(2A=2\left(1+\frac{1}{2}+...+\frac{1}{2^{100}}\right)\)
\(2A=2+1+...+\frac{1}{2^{99}}\)
\(2A-A=\left(2+1+...+\frac{1}{2^{99}}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^{100}}\right)\)
\(A=2-\frac{1}{2^{100}}\)
b)có đứa làm rồi
c)Đặt C=Tổng trên
\(3C=3\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{300}}\right)\)
\(3C=1+\frac{1}{3}+...+\frac{1}{3^{299}}\)
\(3C-C=\left(1+\frac{1}{3}+...+\frac{1}{3^{299}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{300}}\right)\)
\(2C=1-\frac{1}{3^{300}}\)
\(C=\frac{1-\frac{1}{3^{300}}}{2}\)