\(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\)
Những câu hỏi liên quan
frac{1}{2^2}+frac{1}{4^2}+frac{1}{6^2}+..+frac{1}{100^2}frac{1}{2^2}left(1+frac{1}{2^2}+frac{1}{3^2}+...+frac{1}{50^2}right)Có frac{1}{2^2} frac{1}{1.2}1-frac{1}{2} frac{1}{3^2} frac{1}{2.3}frac{1}{2}-frac{1}{3}....v........v............ frac{1}{50^2} frac{1}{49.50}frac{1}{49}-frac{1}{50}Cộng lại 1+frac{1}{2^2}+frac{1}{3^2}+...+frac{1}{50^2} 1+1-frac{1}{2}+frac{1}{2}-frac{1}{3}+...+frac{1}{49}-frac{1}{50}2-frac{1}{50}Rightarrow VT frac{1}{2^2}left(2-frac{1}{50}right)frac{1}{2...
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\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+..+\frac{1}{100^2}=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
Có \(\frac{1}{2^2}< \frac{1}{1.2}=1-\frac{1}{2}\) \(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)....v........v............ \(\frac{1}{50^2}< \frac{1}{49.50}=\frac{1}{49}-\frac{1}{50}\)
Cộng lại \(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=2-\frac{1}{50}\)
\(\Rightarrow VT< \frac{1}{2^2}\left(2-\frac{1}{50}\right)=\frac{1}{2}-\frac{1}{2^2.50}< \frac{1}{2}\left(Đpcm\right)\)
![❄Đờ[̲̅i̲̅]❤Là❤Bể❤K[̲̅h̲̅...](https://hoc24.vn/images/avt/avt2782456_256by256.jpg)
ủa toán lớp mấy chứ ko phải lớp 1
uk ko phải toán lớp 1
Tính giá trị của :
D=\(\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2019^2}\right)x\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\right)-\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2020^2}\right)x\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}\right)\)
Đặt \(a=\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{2019^2}\)
\(b=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2020^2}\)
Khi đó : \(D=ab-\left(b+1\right)\left(a-1\right)\)
\(\Rightarrow D=ab-\left(ab+a-b-1\right)\)
\(\Rightarrow D=b-a+1=\frac{1}{2020^2}-1+1=\frac{1}{2020^2}\)
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Tìm x biết :
\(\frac{1+\frac{1+\frac{1+\frac{3}{2}}{2}}{2}}{1+\frac{2}{1+\frac{2}{1+\frac{2}{3}}}}-x=\frac{\frac{\frac{\frac{2}{3}+1}{3}+1}{3}+1}{\frac{3}{\frac{3}{\frac{3}{2}+1}+1}+1}\)
\(1+\frac{1+\frac{1+\frac{3}{2}}{2}}{2}=1+\frac{1+\frac{\frac{5}{2}}{2}}{2}=1+\frac{1+\frac{5}{4}}{2}=1+\frac{\frac{9}{4}}{2}=1+\frac{9}{8}=\frac{17}{8}\)
\(1+\frac{2}{1+\frac{2}{1+\frac{2}{3}}}=1+\frac{2}{1+\frac{2}{\frac{5}{3}}}=1+\frac{2}{1+\frac{6}{5}}=1+\frac{2}{\frac{11}{5}}=1+\frac{10}{11}=\frac{21}{11}\)
\(1+\frac{1+\frac{1+\frac{2}{3}}{3}}{3}=1+\frac{1+\frac{\frac{5}{3}}{3}}{3}=1+\frac{1+\frac{5}{9}}{3}=1+\frac{\frac{14}{9}}{3}=1+\frac{14}{27}=\frac{41}{27}\)
\(\frac{3}{\frac{3}{\frac{3}{\frac{3}{2}+1}+1}+1}+1=1+\frac{3}{\frac{3}{\frac{3}{\frac{5}{2}}+1}+1}=1+\frac{3}{\frac{3}{\frac{6}{5}+1}+1}=1+\frac{3}{\frac{15}{11}+1}=\frac{59}{26}\)
suy ra
\(\frac{\frac{17}{18}}{\frac{21}{11}}-x=\frac{187}{378}-x=\frac{\frac{41}{27}}{\frac{59}{26}}=\frac{1066}{1593}\Rightarrow x=-\frac{1297}{7434}\)
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\(\frac{1+\frac{1+\frac{5}{4}}{2}}{1+\frac{2}{1+\frac{6}{5}}}-x=\frac{\frac{\frac{\frac{5}{3}}{3}+1}{3}+1}{\frac{3}{\frac{\frac{3}{2}+1}{\frac{5}{2}}}+1}\)
\(\frac{1+\frac{9}{\frac{4}{2}}}{1+\frac{2}{\frac{11}{5}}}-x=\frac{\frac{\frac{14}{9}}{3}+1}{\frac{3}{1}+1}\)
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\(M=\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{2020^2}\right)X\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\right)-\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2020^2}\right)X\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\right)\)Làm nhanh và ngắn gọn nhất có thể nhé ! mình tik cho 10 tik
\(M=\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2020^2}\right)\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\right)-\left(\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\right)\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\right)\)
\(M=\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2020^2}\right)\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\right)(1-1)\)
\(M=\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2020^2}\right)\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\right).0\)
\(M=0\)
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Vì số bị trừ và số trừ gồm hai tích đảo ngược nhau nên M=0
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left(frac{-5}{12}+frac{7}{4}-frac{3}{8}right)-left[4frac{1}{2}-7frac{1}{3}right]-left(frac{1}{4}-frac{5}{2}right)left[2frac{1}{4}-5frac{3}{2}right]-left(frac{3}{10}-1right)-5frac{1}{2}+left(frac{1}{3}-frac{5}{6}right)frac{4}{7}-left(3frac{2}{5}-1frac{1}{2}right)-frac{5}{21}+left[3frac{1}{2}-4frac{2}{3}right]frac{1}{8}-1frac{3}{4}+left(frac{7}{8}-3frac{7}{2}+frac{3}{4}right)-left[frac{7}{4}-frac{5}{8}right]left(frac{3}{5}-2frac{1}{10}+frac{11}{20}right)-left[frac{-3}{4}+1frac{7}{2}right]left[-2fr...
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\(\left(\frac{-5}{12}+\frac{7}{4}-\frac{3}{8}\right)-\left[4\frac{1}{2}-7\frac{1}{3}\right]-\left(\frac{1}{4}-\frac{5}{2}\right)\)
\(\left[2\frac{1}{4}-5\frac{3}{2}\right]-\left(\frac{3}{10}-1\right)-5\frac{1}{2}+\left(\frac{1}{3}-\frac{5}{6}\right)\)
\(\frac{4}{7}-\left(3\frac{2}{5}-1\frac{1}{2}\right)-\frac{5}{21}+\left[3\frac{1}{2}-4\frac{2}{3}\right]\)
\(\frac{1}{8}-1\frac{3}{4}+\left(\frac{7}{8}-3\frac{7}{2}+\frac{3}{4}\right)-\left[\frac{7}{4}-\frac{5}{8}\right]\)
\(\left(\frac{3}{5}-2\frac{1}{10}+\frac{11}{20}\right)-\left[\frac{-3}{4}+1\frac{7}{2}\right]\)
\(\left[-2\frac{1}{5}-2\frac{2}{3}\right]-\left(\frac{1}{15}-5\frac{1}{2}\right)+\left[\frac{-1}{6}+\frac{1}{3}\right]\)
\(1\frac{1}{8}-\left(\frac{1}{15}-\frac{1}{2}+\frac{-1}{6}\right)+\left[\frac{5}{4}+\frac{3}{2}\right]\)
\(\frac{5}{6}-\left(1\frac{1}{3}-1\frac{1}{2}\right)+\left[\frac{5}{12}-\frac{3}{4}-\frac{1}{6}\right]\)
\(1\frac{1}{4}-\left(\frac{7}{12}-\frac{2}{3}-1\frac{3}{8}\right)+\left[\frac{5}{24}-2\frac{1}{2}\right]-\frac{1}{6}-\left[\frac{-3}{4}\right]\)
\(-2\frac{1}{5}+2\frac{3}{10}-\left(\frac{6}{20}-\left[\frac{2}{8}-1\frac{1}{2}\right]\right)+\left[\frac{7}{20}-1\frac{1}{4}\right]\)
\(-\left[1\frac{2}{3}-3\frac{1}{2}+\frac{1}{4}\right]+\left(\frac{2}{6}-\frac{5}{12}\right)-\left(\frac{1}{3}-\left[\frac{1}{4}-\frac{1}{3}\right]\right)\)
\(-\frac{4}{5}-\left(1\frac{1}{10}-\frac{7}{10}\right)+\left[\frac{3}{4}-1\frac{1}{5}\right]+1\frac{1}{2}\)
\(\frac{3}{21}-\frac{5}{14}+\left[1\frac{1}{3}-5\frac{1}{2}+\frac{5}{14}\right]-\left(\frac{1}{6}-\frac{3}{7}+\frac{1}{3}\right)\)
\(-1\frac{2}{5}+\left[1\frac{3}{10}-\frac{7}{20}-1\frac{1}{4}\right]-\left(\frac{1}{5}-\left[\frac{3}{4}-1\frac{1}{2}\right]\right)\)
\(2\frac{1}{3}-\left(\frac{1}{2}-2\frac{1}{6}+\frac{3}{4}\right)+\left[\frac{5}{12}-1\frac{1}{3}\right]-\frac{7}{8}+3\frac{1}{2}\)
\(2\frac{1}{4}-1\frac{3}{5}-\left(\frac{9}{20}-\frac{7}{10}\right)+\left[1\frac{3}{5}-2\frac{1}{2}\right]+\frac{3}{4}\)
\(\left[\frac{8}{3}-5\frac{1}{4}+\frac{1}{6}\right]-\frac{7}{4}+\frac{-5}{12}-\left(1-1\frac{1}{2}+\frac{1}{3}\right)\)
\(\left(\frac{1}{4}-\left[1\frac{1}{4}-\frac{7}{10}\right]+\frac{1}{2}\right)-2\frac{1}{5}-1\frac{3}{10}+\left[1-\frac{1}{2}\right]\)
TRÌNH BÀY GIÚP MÌNH NHA
Tính:
\(A=\sqrt{\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{4^2}}+\sqrt{\frac{1}{1^2}+\frac{1}{4^2}+\frac{1}{5^2}}+...+\sqrt{\frac{1}{1^2}+\frac{1}{2012^2}+\frac{1}{2013^2}}\)
Xét \(P=\sqrt{\frac{1}{1^2}+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}\) với a>0
\(P^2=1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}\)
\(=\frac{a^2\left(a+1\right)^2+\left(a+1\right)^2+a^2}{a^2\left(a+1\right)^2}\)
\(=\frac{a^2\left(a^2+2a+1+1\right)+\left(a+1\right)^2}{a^2\left(a+1\right)^2}\)
\(=\frac{a^4+2a^2\left(a+1\right)+\left(a+1\right)^2}{a^2\left(a+1\right)^2}\)
\(=\frac{\left(a^2+a+1\right)^2}{a^2\left(a+1\right)^2}\)
\(=\left(\frac{a^2+a+1}{a\left(a+1\right)}\right)^2\)
Do a>o nên \(P=\frac{a^2+a+1}{a\left(a+1\right)}=1+\frac{1}{a}-\frac{1}{a+1}\)
Áp dụng kết quả của P ta có:
\(A=\left(1+\frac{1}{1}-\frac{1}{2}\right)+\left(1+\frac{1}{2}+\frac{1}{3}\right)+....+\left(1+\frac{1}{2012}-\frac{1}{2013}\right)\) \(A=2012+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.....+\frac{1}{2012}-\frac{1}{2013}\right)\)
\(A=2012+1-\frac{1}{2013}\)
\(A=2013-\frac{1}{2013}=\frac{4052168}{2013}\)
Vậy \(A=\frac{4052168}{2013}\)
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Tính nhah ---- giúp mik giải nâ các bn thank nhiều nhiềua)frac{1+frac{1}{2}+frac{1}{3}+frac{1}{4}}{1-frac{1}{2}+frac{1}{3}-frac{1}{4}}:frac{3+frac{3}{2}+frac{3}{3}+frac{3}{4}}{2-frac{2}{2}+frac{2}{3}-frac{2}{4}}+frac{1}{3}b) frac{frac{1}{3}-frac{1}{5}-frac{1}{7}}{frac{2}{3}-0,4-frac{2}{7}}+frac{frac{3}{8}-frac{3}{16}-frac{3}{32}+frac{3}{64}}{frac{1}{4}-frac{1}{8}-frac{1}{16}+frac{1}{32}}c) frac{0,4-frac{2}{9}+frac{2}{11}}{1,4-frac{7}{9}+frac{7}{11}}-frac{frac{1}{3}-0,25+frac{1}{5}}{1frac{1}{6}-0...
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Tính nhah ---- giúp mik giải nâ các bn thank nhiều nhiều
a)\(\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}}:\frac{3+\frac{3}{2}+\frac{3}{3}+\frac{3}{4}}{2-\frac{2}{2}+\frac{2}{3}-\frac{2}{4}}+\frac{1}{3}\)
b) \(\frac{\frac{1}{3}-\frac{1}{5}-\frac{1}{7}}{\frac{2}{3}-0,4-\frac{2}{7}}+\frac{\frac{3}{8}-\frac{3}{16}-\frac{3}{32}+\frac{3}{64}}{\frac{1}{4}-\frac{1}{8}-\frac{1}{16}+\frac{1}{32}}\)
c) \(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\)
frac{frac{1}{2}+frac{1}{3}+......+frac{1}{2013}}{frac{2012}{1}+frac{2012}{2}+frac{2011}{3}+...+frac{1}{2013}}frac{frac{1}{2}+frac{1}{3}+..+frac{1}{2013}}{frac{2012}{1}+2+frac{2012}{2}+1+frac{2011}{3}+1+...+frac{1}{2013}+1-2014}frac{frac{1}{2}+frac{1}{3}+...+frac{1}{2013}}{frac{2014}{1}+frac{2014}{2}+...+frac{2014}{2013}-2014}frac{frac{1}{2}+frac{1}{3}+...+frac{1}{2013}}{2014left(1+frac{1}{2}+frac{1}{3}+...+frac{1}{2013}-1right)}frac{1}{2014}
Đọc tiếp
\(\frac{\frac{1}{2}+\frac{1}{3}+......+\frac{1}{2013}}{\frac{2012}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)
=\(\frac{\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2013}}{\frac{2012}{1}+2+\frac{2012}{2}+1+\frac{2011}{3}+1+...+\frac{1}{2013}+1-2014}\)
=\(\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}{\frac{2014}{1}+\frac{2014}{2}+...+\frac{2014}{2013}-2014}\)
=\(\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}{2014\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}-1\right)}\)
=\(\frac{1}{2014}\)
Chứng minh frac{1}{5}+frac{1}{13}+frac{1}{14}+frac{1}{15}+frac{1}{61}+frac{1}{62}+frac{1}{63} frac{1}{2}frac{1}{2}b,frac{1}{^{2^2}}+frac{1}{4^2}+frac{1}{6^2}+...+frac{1}{100^2} frac{1}{2}c,frac{3}{4}+frac{8}{9}+frac{15}{16}+...+frac{2499}{2500}48d,frac{1}{6} frac{1}{5^2}+frac{1}{6^2}+frac{1}{7^2}+...+frac{1}{100^2} frac{1}{4}
Đọc tiếp
Chứng minh \(\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{2}\)\(\frac{1}{2}\)
b,\(\frac{1}{^{2^2}}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
c,\(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}>48\)
d,\(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\)
a, \(\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)\)
Ta có: \(\frac{1}{13}< \frac{1}{12};\frac{1}{14}< \frac{1}{12};\frac{1}{15}< \frac{1}{12}\Rightarrow\frac{1}{13}+\frac{1}{14}+\frac{1}{15}< \frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{3}{12}=\frac{1}{4}\)
\(\frac{1}{61}< \frac{1}{60};\frac{1}{62}< \frac{1}{60};\frac{1}{63}< \frac{1}{60}\Rightarrow\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{60}+\frac{1}{60}+\frac{1}{60}=\frac{3}{60}=\frac{1}{20}\)
\(\Rightarrow\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{1}{2}\)
Vậy...
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b, Đặt A là tên của tổng trên
Ta có: \(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\right)\)
Đặt B là biêu thức trong ngoặc
Ta có: \(1=1;\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};....;\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow B< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(\Rightarrow B< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow B< 2-\frac{1}{50}< 2\)
Thay B vào A ta được:
\(A< \frac{1}{2^2}.2=\frac{1}{2}\)
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c, Đặt C = \(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}\)
\(C=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{2500}\right)\)
\(C=\left(1+1+1+....+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{2500}\right)\)
\(C=49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)\)
Đặt D là biểu thức trong ngoặc
Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};....;\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow D< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}< 1\)
\(\Rightarrow-D>-1\)
=>\(C=49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)>49-1=48\)
Vậy C > 48
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Xem thêm câu trả lời
Tính nhanh :
1)A=\(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-...+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)
2)B=\(\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+...+\frac{1}{2^{99}}\)
3)C=\(\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-\frac{1}{2^{10}}+...-\frac{1}{2^{58}}\)
\(2A=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+.....-\frac{1}{2^{99}}\Rightarrow2A+A=3A=\left(1-\frac{1}{2}+\frac{1}{2^2}-....-\frac{1}{2^{99}}\right)+\left(\frac{1}{2}-\frac{1}{2^2}+......-\frac{1}{2^{100}}\right)=1-\frac{1}{2^{100}}=\frac{2^{100}-1}{2^{100}}\Rightarrow A=\frac{2^{100}-1}{3.2^{100}}\)
\(2,4B=2+\frac{1}{2}+\frac{1}{2^3}+.....+\frac{1}{2^{97}}\Rightarrow4B-B=3B=\left(2+\frac{1}{2}+....+\frac{1}{2^{97}}\right)-\left(\frac{1}{2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)=2-\frac{1}{2^{99}}=\frac{2^{100}-1}{2^{99}}\Rightarrow B=\frac{2^{100}-1}{3.2^{99}}\)
\(3,C=\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-....-\frac{1}{2^{58}}\Rightarrow8C=4-\frac{1}{2}+\frac{1}{2^4}-.....-\frac{1}{2^{55}}\Rightarrow8C+C=9C=\left(4-\frac{1}{2}+\frac{1}{2^4}-....-\frac{1}{2^{55}}\right)+\left(\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-....-\frac{1}{2^{58}}\right)=4-\frac{1}{2^{58}}=\frac{2^{60}-1}{2^{58}}\Rightarrow C=\frac{2^{60}-1}{9.2^{58}}\)
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