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XO
8 tháng 7 2021
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
16 tháng 5 2018
\(S=1+2\cdot3+3\cdot3^2+4\cdot3^3+...+101\cdot3^{100}=\left(1+3+3^2+...+3^{100}\right)+\left(3+3^2+...+3^{100}\right)+...+3^{100}\)\(S=\left(1+...+3^{100}\right)+3\left(1+...+3^{99}\right)+3^2\left(1+...+3^{98}\right)+...+3^{100}\)
\(S=1\cdot A_{100}+3\cdot A_{99}+3^2\cdot A_{98}+...+3^{100}\)
\(A_i=1+3+3^2+...+3^i=\frac{3^{i+1}-1}{2}\)
\(S=\frac{3^{101}-1}{2}+\frac{3\left(3^{100}-1\right)}{2}+\frac{3^2\left(3^{99}-1\right)}{2}+...+\frac{3^{100}\left(3-1\right)}{2}\)
\(2S=3^{101}\cdot101-\left(1+2+3+...+3^{100}\right)=101\cdot3^{101}-A_{100}=101\cdot3^{101}-\frac{3^{101}-1}{2}\)
\(2S=\frac{201\cdot3^{101}+1}{2}\Leftrightarrow S=\frac{201\cdot3^{101}+1}{4}\)
SH
29 tháng 9 2018
((3\(^2\)))\(^2\) - ((-5\(^2\)))\(^2\) + ((-2\(^3\)))\(^2\)
= 81 - 625 + 64
= -544+ 64
= -480
2\(^4\) + 8[(-2)\(^2\) :\(\dfrac{1}{2}\)]\(^0\) - 2\(^{-2}\). 4 + (-2)\(^2\)
= 16+ 8.1 - \(\dfrac{1}{4}\). 4 + 4
= 16+ 8- 1+4
= 27
2\(^4\) + 3(\(\dfrac{1}{2}\))\(^0\) + 2\(^{-2}\).8 + [(-2)\(^3\). \(\dfrac{1}{2^4}\)].2 - \(\dfrac{1}{2}\)
= 16 + 3.1 +\(\dfrac{1}{4}\).8 + [(-8).\(\dfrac{1}{16}\)].2 -\(\dfrac{1}{2}\)
= 16 + 3+ 2 + \(\dfrac{-1}{2}\).2- \(\dfrac{1}{2}\)
= 21 + (-1)- \(\dfrac{1}{2}\)
= 20-\(\dfrac{1}{2}\) = \(\dfrac{40}{2}\) - \(\dfrac{1}{2}\)= \(\dfrac{39}{2}\)
\(\dfrac{15^{10}.5^{10}}{75^{10}}\) + \(\dfrac{\left(0,8\right)^5}{\left(0,4\right)^6}\)
= \(\dfrac{\left(15.5\right)^{10}}{75^{10}}\) + \(\dfrac{\left(0,4.2\right)^5}{\left(0.4\right)^6}\)
= \(\dfrac{75^{10}}{75^{10}}\) + \(\dfrac{\left(0,4\right)^5.2^5}{\left(0,4\right)^6}\)
= 1 + \(\dfrac{2^5}{0,4}\) = 1+ 80 = 81
\(\dfrac{2^{13}.9^4}{6^3.8^3}\)
= \(\dfrac{2^{13}.\left(3^2\right)^4}{\left(2.3\right)^3.\left(2^3\right)^3}\) = \(\dfrac{2^{13}.3^8}{2^3.3^3.2^9}\)
= \(\dfrac{2^4.3^5}{2^3}\) = 2.3\(^5\) = 486
VM
11 tháng 8 2019
Tính:
a) \(\left(2^{-1}+3^{-1}\right):\left(2^{-1}-3^{-1}\right)+\left(2^{-1}.2^0\right):2^3\)
\(=\left(\frac{1}{2}+\frac{1}{3}\right):\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{2}.1\right):8\)
\(=\frac{5}{6}:\frac{1}{6}+\frac{1}{2}:8\)
\(=5+\frac{1}{16}\)
\(=\frac{81}{16}.\)
Chúc bạn học tốt!