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:\(A=1+\frac{1}{2}\times\left(1+2\right)+\frac{1}{3}\times\left(1+2+3\right)+...+\frac{1}{16}\times\left(1+2+3+...+16\right)\)
\(\Rightarrow A=1+\frac{1}{2}\times\frac{2\times3}{2}+\frac{1}{3}\times\frac{3\times4}{2}+...+\frac{1}{16}\times\frac{16\times17}{2}\)
\(\Rightarrow A=1+\frac{3}{2}+\frac{4}{2}+...+\frac{17}{2}\)
\(\Rightarrow A=\frac{1}{2}\times\left(2+3+4+...+17\right)\)
\(\Rightarrow A=\frac{1}{2}\times152=76\)
Vậy A=76
\(3B=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(B=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow4B=3B+B=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
+ Đặt \(M=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
\(3M=3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
\(\Rightarrow4M=3M+M=3-\frac{1}{3^{99}}\)
\(\Rightarrow M=\frac{3}{4}-\frac{1}{3^{99}\cdot4}\)
\(\Rightarrow4B=M-\frac{100}{3^{100}}=\frac{3}{4}-\frac{1}{3^{99}\cdot4}-\frac{100}{3^{100}}\)
\(\Rightarrow B=\frac{3}{16}-\frac{1}{3^{99}\cdot16}-\frac{100}{3^{100}\cdot4}\) \(\Rightarrow B< \frac{3}{16}\)
a) \(2A=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\)
\(A=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\)
\(\Rightarrow3A=2A+A=1-\frac{1}{2^6}\)
\(\Rightarrow A=\frac{1}{3}-\frac{1}{2^6\cdot3}< \frac{1}{3}\) ( đpcm )
\(A=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+...+\frac{1}{16}.\left(1+2+...+16\right)\)
\(=1+\frac{1}{2}.2.3:2+\frac{1}{3}.3.4:2+...+\frac{1}{16}.16.17:2=1+\frac{3}{2}+\frac{4}{2}+...+\frac{17}{2}=\frac{2+3+4+...+17}{2}=\frac{152}{2}=76\)
a) \(A=\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2019}}\)
\(5A=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2018}}\)
\(4A=5A-A=\frac{1}{5}-\frac{1}{5^{2019}}\)
\(A=\frac{1}{20}-\frac{1}{4.5^{2019}}< \frac{1}{20}< \frac{1}{2}\)
b) Đề có sai không mà đằng cuối lại là \(\frac{1}{4^2}\)lặp lại lần nữa.
c) \(C=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)
\(2C=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}\)
\(3C=2C+C=1-\frac{1}{64}< 1\)
\(C< \frac{1}{3}\)
d) Xem lại đề nữa đi e, nếu trừ hai vế cho \(\frac{1}{3}\)thì vế trái > 0 > vế phải rồi
e) \(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}>\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\)(10 số hạng)
\(=\frac{10}{50}=\frac{1}{5}\)
Tương tự: \(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}>\frac{1}{6}\)
\(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{70}>\frac{1}{7}\)
\(\frac{1}{71}+\frac{1}{72}+...+\frac{1}{80}>\frac{1}{8}\)
\(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{80}>\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}=\frac{533}{840}>\frac{490}{840}=\frac{7}{12}\)
Bài 1:
Vì \(\frac{196}{197+198}< \frac{196}{197};\frac{197}{197+198}< \frac{197}{198}\)
Nên A = \(\frac{196}{197}+\frac{197}{198}>\frac{196}{197+198}+\frac{197}{197+198}=\frac{196+197}{197+198}=B\)
Vậy A > B
\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{15}\left(1+2+...+15\right)+\frac{1}{16}\left(1+2+3+...+16\right)\)
\(A=1+\frac{1}{2}\cdot3+\frac{1}{3}\cdot6+\frac{1}{4}\cdot10+...+\frac{1}{15}+\left[\frac{\left(1+15\right)\cdot15}{2}\right]+\frac{1}{16}\cdot\left[\frac{\left(16+1\right).16}{2}\right]\)
\(A=1+\frac{3}{2}+2+\frac{5}{2}+....+\frac{1}{15}\cdot120+\frac{1}{16}\cdot136\)
\(A=1+\frac{3}{2}+2+\frac{5}{2}+...+8+\frac{17}{2}\)
\(A=\left(1+2+...+8\right)+\left(\frac{3}{2}+\frac{5}{2}+...+\frac{17}{2}\right)\)
Đặt \(B=1+2+...+8\)
\(C=\frac{3}{2}+\frac{5}{2}+...+\frac{17}{2}\)
\(B=1+2+...+8\)
\(\text{Ta thấy tổng B là dãy các số hạng liên tiếp từ 1 đến 8 }\)
\(\Rightarrow\text{số số hạng của B là}:\)\(\left(8-1\right)\div1+1=8\left(sh\right)\)
\(\text{Tổng B là }:\)\(\frac{\left(1+8\right)\cdot8}{2}=36\)
\(C=\frac{3}{2}+\frac{5}{2}+...+\frac{17}{2}\)
\(\Rightarrow C=\frac{3+5+...+17}{2}\)
Đặt \(D=3+5+...+17\)
\(\text{số số hạng của D là}:\)\(\left(17-3\right)\div2+1=8\left(sh\right)\)
\(\text{Tổng D là }:\)\(\frac{\left(3+17\right)\cdot8}{2}=80\)
\(\Rightarrow C=\frac{80}{2}=40\)
Thay B và C vào biểu thức A , ta được
\(A=36+40=76\)
Vậy A = 76
\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)\)\(+...+\frac{1}{16}\left(1+2+3+...+16\right)\)
\(\Rightarrow A=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+\frac{1}{4}.\frac{4.5}{2}+...+\frac{1}{16}.\frac{16.17}{2}\)
\(\Rightarrow A=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{17}{2}\)
\(\Rightarrow A=\frac{\frac{17.18}{2}-1}{2}=76.\)
Vậy \(A=76.\)