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đặt \(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow3A=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow A+3A=\left(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\right)+\left(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\right)\)
\(\Rightarrow4A=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)<\(B=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
\(\Rightarrow3B=3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
\(\Rightarrow B+3B=\left(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\right)+\left(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\right)\)
\(\Rightarrow4B=3-\frac{1}{3^{98}}<3\)
\(\Rightarrow B<\frac{3}{4}\Rightarrow4A<\frac{3}{4}\Rightarrow A<\frac{3}{16}\)
\(\RightarrowĐPCM\)
Đặt 1/3_2/3^2+3/3^3_.......+99/3^99_100/3^100<3/16=A
3A=1_2/3+3/3^2_4/3^3+....+99/3^98_100/3^99
Lấy 3A+A=4A=1_1/3+1/3^2_1/3^3+1/3^4_....._1/3^99_100/3^100
4A<1_1/3+1/3^2_1/3^3+1/3^4_...+1/3^98_1/3^99(1)
Đặt B=1_1/3+1/3^2_1/3^3_1/3^4_....+1/3^98_1/3^99
+
3B=2+1/3_1/3^2+1/3^3+......+1/3^97_1/3^98
4B=3_1/3^99<3 suy ra 4B<3 suy ra B<3/4(2)
Từ (1) và (2) suy ra 4A<3/4
Suy ra A<3/16
MỚI LÀM LÚC TỐI,HÊN QUÁ:
\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(2A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(6A=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
\(4A=3-\left(\frac{101}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(4A=3-\frac{203}{3^{100}}\)
\(A=\frac{3}{4}-\frac{203}{3^{100}\cdot4}< \frac{3}{4}\)
Ta có : A = \(\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{99}{5^{100}}\)
=> 5A = \(\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{99}{5^{99}}\)
=> 5A - A = \(\left(\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{99}{5^{99}}\right)-\left(\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{99}{5^{100}}\right)\)
=> 4A \(=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)
=> 20A = \(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{98}}-\frac{99}{5^{99}}\)
Lấy 20A trừ A ta có :
20A - A = \(\left(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{98}}-\frac{99}{5^{99}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{99}}-\frac{99}{5^{100}}\right)\)
16A = \(1-\frac{99}{5^{99}}+\frac{99}{5^{100}}=1+99\left(\frac{1}{5^{100}}-\frac{1}{5^{99}}\right)=1-\frac{99.4}{5^{100}}\)
=> A = \(\frac{1}{16}-\frac{99}{4.5^{100}}< \frac{1}{16}\left(\text{ĐPCM}\right)\)
Ta có :A=\(\frac{1}{5^2}+\frac{2}{5^3}+.....+\frac{99}{5^{100}}\)
5A=\(\frac{1}{5}+\frac{2}{5^2}+.....+\frac{99}{5^{99}}\)
5A -A=\(\left(\frac{1}{5}+\frac{2}{5^2}+...+\frac{99}{5^{99}}\right)\)-\(\left(\frac{1}{5^2}+\frac{2}{5^3}+...+\frac{99}{5^{100}}\right)\)
4A =\(\frac{1}{5}+\frac{1}{5^2}+....+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)
Đặt B=\(\frac{1}{5}+\frac{1}{5^2}+.....+\frac{1}{5^{99}}\)
5B=\(1+\frac{1}{5}+...+\frac{1}{5^{98}}\)
5B - B =\(\left(1+\frac{1}{5}+...+\frac{1}{5^{98}}\right)\)- \(\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\right)\)
4B =\(1-\frac{1}{5^{99}}\)
Ta có :4A = B -\(\frac{99}{5^{100}}\)
16A = 4B -\(\frac{4.99}{5^{100}}\)=\(1-\frac{1}{5^{99}}-\frac{4.99}{5^{100}}\)
A = \(\frac{1}{16}-\frac{1}{5^{99}.16}-\frac{99}{5^{100}.4}\)< \(\frac{1}{16}\)
Suy ra: A <\(\frac{1}{16}\)
Nhầm đầu bài nhoa:
Phải là \(-\frac{100}{3^{100}}\)
đề phía sau dấu < hình như sai
Một số 3/16 thôi mk ghi nhầm