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\(B=1-\frac{1}{2}\left(1+2\right)-\frac{1}{3}.\left(1+2+3\right)-\frac{1}{4}.\left(1+2+3+4\right)-...-\frac{1}{20}.\left(1+2+3+...+20\right)\)
\(B=1-\frac{1}{2}.\left(1+2\right).2:2-\frac{1}{4}.\left(1+4\right).4:2-...-\frac{1}{20}.\left(1+20\right).20:2\)
\(B=1-3:2-5:2-...-21:2\)
\(B=1-3.\frac{1}{2}-5.\frac{1}{2}-...-21.\frac{1}{2}\)
\(B=1-\frac{1}{2}.\left(3+5+...+21\right)\)
Đặt C = 3 + 5 + ... + 21
Số số hạng của tổng C là: (21 - 3) : 2 + 1 = 10 (số)
=> C = (3 + 21) x 10 : 2 = 24 x 5 = 120
=> \(A=1-\frac{1}{2}.120\)
\(A=1-60=-59\)
Huong dẫn: \(1+2+...+n=\frac{n\left(n+1\right)}{2}\) ( n\(\in\)N*) áp dụng vào từng cái ngoặc
\(S=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+.......+\frac{1}{20}\left(1+2+3+......+20\right)\)
\(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+......+\frac{1}{20}.\frac{20.21}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+......+\frac{21}{2}=\frac{2+3+4+.....+21}{2}=\frac{20.23}{2}=230\)
Bài 1:
a) \(x-\frac{20}{11.13}-\frac{20}{13.15}-...-\frac{20}{53.55}=\frac{3}{11}\)
\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)
\(x-\frac{20}{2}.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-10.\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-10\cdot\frac{4}{55}=\frac{3}{11}\)
\(x-\frac{8}{11}=\frac{3}{11}\)
\(x=1\)
b) \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
\(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
\(\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
\(2.\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(2.\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{18}\)
=> x + 1 =18
x = 17
bài 2 ko bk lm, xl nha
Đặt \(A=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{20}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{20}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{19}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{19}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{20}}\right)\)
\(A=1-\frac{1}{2^{20}}\)
\(A=\frac{2^{20}}{2^{20}}-\frac{1}{2^{20}}\)
\(A=\frac{2^{20}-1}{2^{20}}\)
Vậy chọn câu a)
\(=1+\frac{1}{2}.\frac{\left(1+2\right).3}{2}+\frac{1}{3}.\frac{\left(1+3\right).3}{2}+...+\frac{1}{20}.\frac{\left(1+20\right).20}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+...+\frac{21}{2}\)
\(=\frac{2+3+4+...+21}{2}=\frac{\left(21+2\right).20}{2}=23.10=230\)
Với 20 là số số hạng của dãy 2, 3, 4, ...., 21