\(\frac{1}{2}\)\(+\)\(\frac{3}{2^2}\)\(+\)\(\frac{7}{2^3}\)\(+\)... \(+\)\(\frac{2^{100}-1}{2^{100}}\) Chứng minh B > 99
Mình đag cần rất gấp. Ai làm nhanh mình tick . Mọi ng giúp mình với
\(A=(\frac{1}{7^2}+\frac{1}{7^4}+\frac{1}{7^6}+...+\frac{1}{7^{100}})\div(1-\frac{1}{7^{100}})\)Mình đag cần rất gấp . Mong mn giúp mình với
Ai làm nhanh mình tick
Đặt S = \(\frac{1}{7^2}+\frac{1}{7^4}+\frac{1}{7^6}+...+\frac{1}{7^{100}}\)
=> 72S = 49S = \(1+\frac{1}{7^2}+\frac{1}{7^4}+...+\frac{1}{7^{98}}\)
=> 49S - S = \(\left(1+\frac{1}{7^2}+\frac{1}{7^4}+...+\frac{1}{7^{98}}\right)-\left(\frac{1}{7^2}+\frac{1}{7^4}+\frac{1}{7^6}+...+\frac{1}{7^{100}}\right)\)
=> 48S = \(1-\frac{1}{7^{100}}\)
=> \(S=\frac{1-\frac{1}{7^{100}}}{48}\)
Khi đó A = \(\left(\frac{1-\frac{1}{7^{100}}}{48}\right):\left(1-\frac{1}{7^{100}}\right)=\frac{1}{48}\)
\(\frac{3}{2}\)\(+\)\(\frac{5}{2^2}\)\(+\)\(\frac{9}{2^3}\)\(+\)... \(+\)\(\frac{3^{100}+1}{2^{100}}\)Chứng minh 100 < B < 101
Mình đag cần gấp. Ai lm nhanh mình tick . Các bạn giúp mình với
Sửa đề \(\frac{3}{2}+\frac{5}{2^2}+\frac{9}{2^3}+...+\frac{2^{100}+1}{2^{100}}=\frac{2+1}{2}+\frac{2^2+1}{2^2}+\frac{2^3+1}{2^3}+...+\frac{2^{100}+1}{2^{100}}\)
= \(\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)(100 hạng tử 1)
= \(100+\left(1-\frac{1}{2^{100}}\right)=101-\frac{1}{2^{100}}< 101\)(1)
Vì \(-\frac{1}{2^{100}}>-1\Rightarrow101-\frac{1}{2^{100}}>101-1\Rightarrow B>100\)(2)
Từ (1) và (2) => 100 < B < 101
Tìm các số nguyên x, biết: \(-4\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le\frac{-2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
Mình đag cần rất gấp. Ai làm nhanh mình tick. Mong mn giúp mình với
\(\frac{2}{1.4.7}+\frac{2}{4.7.10}+...+\frac{2}{58.61.64}\)Mình đag cần rất gấp. Mọi người giúp mình với. Ai lm nhanh mình tick
\(\frac{2}{1.4.7}+\frac{2}{4.7.10}+...+\frac{2}{58.61.64}\)
\(=\frac{1}{3}\left(\frac{6}{1.4.7}+\frac{6}{4.7.10}+...+\frac{6}{58.61.64}\right)\)
\(=\frac{1}{3}\left(\frac{7-1}{1.4.7}+\frac{10-4}{4.7.10}+...+\frac{64-58}{58.61.64}\right)\)
\(=\frac{1}{3}\left(\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+...+\frac{1}{58.61}-\frac{1}{61.64}\right)\)
\(=\frac{1}{3}\left(\frac{1}{1.4}-\frac{1}{61.64}\right)\)
\(=\frac{325}{3904}\)
\(\frac{1}{2}\)\(+\)\(\frac{3}{2^2}\)\(+\)\(\frac{7}{2^3}\)\(+\)... \(+\)\(\frac{2^{100}-1}{2^{100}}\)Chứng minh B > 99
Mình đag cần cần rất gấp. mn giúp mình với
Ta có \(\frac{1}{2}+\frac{3}{2^2}+\frac{7}{2^3}+...+\frac{2^{100}-1}{2^{100}}\)
= \(\frac{2-1}{2}+\frac{2^2-1}{2^2}+\frac{2^3-1}{2^3}+...+\frac{2^{100}-1}{2^{100}}\)
= \(1-\frac{1}{2}+1-\frac{1}{2^2}+1-\frac{1}{2^3}+...+1-\frac{1}{2^{100}}\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)(100 hạng tử 1)
\(=100-\left(1-\frac{1}{2^{100}}\right)=100-1+\frac{1}{2^{100}}=99+\frac{1}{2^{100}}>99\)(đpcm)
\(A=(\frac{1}{2}+\frac{1}{2^5}+\frac{1}{2^9}+...+\frac{1}{2^{101}})\div(2^3-\frac{1}{2^{101}})\)Mình đag cần rất gấp. Tí nx mình phải nộp. Ai lm nhanh mình tick. Mong mn giúp mình
Đặt S = \(\frac{1}{2}+\frac{1}{2^5}+\frac{1}{2^9}+...+\frac{1}{2^{101}}\)
=> 24S = 16S = \(2^3+\frac{1}{2}+\frac{1}{2^5}+...+\frac{1}{2^{97}}\)
=> 16S - S = \(2^3+\frac{1}{2}+\frac{1}{2^5}+...+\frac{1}{2^{97}}-\left(\frac{1}{2}+\frac{1}{2^5}+\frac{1}{2^9}+...+\frac{1}{2^{101}}\right)\)
=> 15S = \(2^3-\frac{1}{2^{101}}\)
=> S = \(\frac{2^3-\frac{1}{2^{101}}}{15}\)
Khi đó A = \(\frac{2^3-\frac{1}{2^{101}}}{15}:\left(2^3-\frac{1}{2^{101}}\right)=\frac{1}{15}\)
kết bạn đi toán lớp mấy vậy
Tìm các số nguyên x biết : \(-4\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
Mình đag cần rất gấp. Ai lm nhanh mình tick. Mong mn giúp mình với
\(-4\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
\(\Rightarrow-\frac{13}{3}.\left(\frac{3}{6}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{4}{12}-\frac{6}{12}-\frac{9}{12}\right)\)
\(\Rightarrow-\frac{13}{3}.\frac{2}{6}\le x\le-\frac{2}{3}.\frac{-11}{12}\)
\(\Rightarrow\frac{-13}{9}\le x\le\frac{11}{18}\)
\(\Rightarrow\frac{-26}{18}\le x\le\frac{11}{18}\)
=> -1,44444444444........... ≤ x ≤ 0,6111111111...........
Mà x ∈ Z
=> x ∈ { -1 ; 0 }
\(x\in\varnothing\)
1.Chứng minh rằng: \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^3.4^2}+...+\frac{19}{9^2.10^2}< 1\)
2.Chứng minh rằng: \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}< \frac{3}{4}\)
Làm nhanh giúp mình nhé mọi người !!!
Bài 1:
Ta có:
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)
\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{81}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Mà \(\frac{99}{100}< 1\)
\(\Rightarrow\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\left(đpcm\right)\)
Có phải ở sách NCPT ko bn
Bài 2: Đặt \(B=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\)
\(3B=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)
\(3B-B=\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\right)\)
\(2B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(6B=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(6B-2B=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(4B=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)
\(4B=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)
\(4B=3-\frac{303}{3^{100}}+\frac{100}{3^{100}}\)
\(4B=3-\frac{203}{3^{100}}< 3\)
\(B< \frac{3}{4}\left(đpcm\right)\)
\(A\)\(=\)\((1-\frac{1}{6^{100}})\div(\frac{1}{6}+\frac{1}{6^2}+\frac{1}{6^3}+...+\frac{1}{6^{100}})\)Mình đag cần rất gấp . Các bạn giúp mình với
Đặt S = \(\frac{1}{6}+\frac{1}{6^2}+\frac{1}{6^3}+...+\frac{1}{6^{100}}\)
=> 6S = \(1+\frac{1}{6}+\frac{1}{6^2}+...+\frac{1}{6^{99}}\)
=> 6S - S = \(\left(1+\frac{1}{6}+\frac{1}{6^2}+\frac{1}{6^3}+...+\frac{1}{6^{99}}\right)-\left(\frac{1}{6}+\frac{1}{6^2}+\frac{1}{6^3}+...+\frac{1}{6^{100}}\right)\)
=> \(5S=1-\frac{1}{6^{100}}\)
=> \(S=\frac{1-\frac{1}{6^{100}}}{5}\)
Khi đó A = \(\left(1-\frac{1}{6^{100}}\right):\left(\frac{1-\frac{1}{6^{100}}}{5}\right)=5\)