1+1/2*(1+2)+1/3*(1+2+3)+1/4*(1+2+3+4)+....+1/30*(1+2+3+4+...+30)
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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
22 tháng 3 2024
Câu 11:
(\(\dfrac{11}{4}\). \(\dfrac{-5}{9}\) - \(\dfrac{4}{9}\).\(\dfrac{11}{4}\)).\(\dfrac{8}{33}\)
= \(\dfrac{11}{4}\).(\(\dfrac{-5}{9}\) - \(\dfrac{4}{9}\)). \(\dfrac{8}{33}\)
= \(\dfrac{11}{4}\).(-1).\(\dfrac{8}{33}\)
= - \(\dfrac{2}{3}\)
16 tháng 11 2020
B=1+1/2*(1+2)*2/2+1/3*(1+3)*3/2+....+1/30*(1+30)*30/2
=1+1+2/2+1+3/2+...+1+30/2
=2+3+4+...+31/2
=(31+2)*30/2 phần 2
=) B=247,5
3 tháng 2 2022
a) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)
=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)
=\(1-\dfrac{1}{6}\)=\(\dfrac{5}{6}\)
b) \(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
=\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)
=\(\dfrac{1.2}{3.5.2}+\dfrac{1.2}{5.7.2}+\dfrac{1.2}{7.9.2}+\dfrac{1.2}{9.11.2}+\dfrac{1.2}{11.13.2}\)
=\(\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\right)\).
=\(\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\right)\)
=\(\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)=\(\dfrac{1}{2}.\dfrac{10}{39}\)=\(\dfrac{5}{39}\).
c) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)
=\(1-\dfrac{1}{8}=\dfrac{7}{8}\).
d) \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}\)
=\(\dfrac{2^4}{2^5}+\dfrac{2^3}{2^5}+\dfrac{2^2}{2^5}+\dfrac{2}{2^5}+\dfrac{1}{2^5}\)
=\(\dfrac{2^4+2^3+2^2+2+1}{2^5}\)=\(\dfrac{2^5-1}{2^5}=\dfrac{31}{32}\).
e) \(\dfrac{1}{7}+\dfrac{1}{7^2}+\dfrac{1}{7^3}+...+\dfrac{1}{7^{100}}=\dfrac{7^{99}+7^{98}+7^{97}+...+7+1}{7^{100}}=\dfrac{\dfrac{7^{100}-1}{6}}{7^{100}}=\dfrac{7^{100}-1}{6.7^{100}}\)
16 tháng 10 2021
\(1,\\ a,\Leftrightarrow x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\\ b,\Leftrightarrow\left[{}\begin{matrix}x-4=4\\x-4=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=0\end{matrix}\right.\\ c,\Leftrightarrow2x+1=-2\Leftrightarrow x=-\dfrac{3}{2}\\ 2,\\ a,=1\\ b,=\left(\dfrac{13}{4}\right)^2=\dfrac{169}{16}\\ c,=\left(-\dfrac{7}{4}\right)^2=\dfrac{49}{16}\\ d,=\left(\dfrac{3}{7}\right)^{20}:\left(\dfrac{3}{7}\right)^{12}=\left(\dfrac{3}{7}\right)^8=...\\ e,=\left(3\cdot5\cdot\dfrac{2}{3}\right)^2=10^2=100\)
NP
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
11 tháng 9 2023
A = 1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4)+...+ (1 + 2 + 3+...+ 30)
A = 1 + (2+1)\(\times\)2: 2+ (3+1)\(\times\)3: 2 + ....+ (30 +1) \(\times\) 30: 2
A = 1 + 2 \(\times\)3: 2 + 3 \(\times\) 4 : 2 +...+ 30 \(\times\) 31:2
A = \(\dfrac{2+2\times3+3\times4+...+30\times31}{2}\)
Đặt B = 2 + 2 \(\times\) 3 + 3 \(\times\) 4 +...+ 30 \(\times\) 31
1 x 2 x 3 = 1 x 2 x3 = 1 x 2 x 3
2 x 3 x 3 = 2 x 3 x ( 4 - 1) = 2 x 3 x 4 - 1 x 2 x 3
3 x 4 x 3 = 3 x 4 x (5- 2) = 3 x 4 x 5 - 2 x 3 x 4
...........................................................................
30 x 31 x 3 = 30 x 31 x (32 - 29) = 30 x 31 x 32 - 29 x 31 x 30
Cộng vế với vế ta có:
B x 3 = 30 x 31 x 32 ⇒ B = 30 x 31 x 32 : 3 = 14880
A = B : 2 = 14880 : 2 = 7440
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
8 tháng 4 2023
A = 1 + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\) +.......+\(\dfrac{1}{3^{n-1}}\) + \(\dfrac{1}{3^n}\)
3\(\times\) A = 3 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\)+........+ \(\dfrac{1}{3^{n-1}}\)
3A - A = 3 + \(\dfrac{1}{3}\) - 1 - \(\dfrac{1}{3^n}\)
2A = \(\dfrac{7}{3}\) - \(\dfrac{1}{3^n}\)
A = ( \(\dfrac{7}{3}\) - \(\dfrac{1}{3^n}\)): 2
A = \(\dfrac{7.3^{n-1}-1}{3^n}\) : 2
A = \(\dfrac{7.3^{n-1}-1}{2.3^n}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
8 tháng 4 2023
B = \(\dfrac{1}{2}\) - \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\) - \(\dfrac{1}{2^4}\)+......+\(\dfrac{1}{2^{99}}\) - \(\dfrac{1}{2^{100}}\)
2B = 2 - \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) - \(\dfrac{1}{2^3}\)+ \(\dfrac{1}{2^4}\)-.......-\(\dfrac{1}{2^{99}}\)
2B + B = 2 - \(\dfrac{1}{2^{100}}\)
3B = 2 - \(\dfrac{1}{2^{100}}\)
B = ( 2 - \(\dfrac{1}{2^{100}}\)): 3
B = \(\dfrac{2.2^{100}-1}{2^{100}}\) : 3
B = \(\dfrac{2^{101}-1}{3.2^{100}}\)
Đặt \(A=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+............+\frac{1}{30}.\left(1+2+3+..........+30\right)\)
\(\Rightarrow A=1+\frac{1+2}{2}+\frac{1+2+3}{3}+.........+\frac{1+2+3+.........+30}{30}\)
\(\Rightarrow A=1+\frac{\frac{\left(1+2\right).2}{2}}{2}+\frac{\frac{\left(1+3\right).3}{2}}{3}+..............+\frac{\frac{\left(1+30\right).30}{2}}{30}\)
\(\Rightarrow A=1+\frac{\left(1+2\right).2}{2.2}+\frac{\left(1+3\right).3}{2.3}+..............+\frac{\left(1+30\right).30}{2.30}\)
\(\Rightarrow A=1+\frac{1+2}{2}+\frac{1+3}{2}+............+\frac{1+30}{2}\)
\(\Rightarrow A=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+.................+\frac{31}{2}\)
\(\Rightarrow A=\left(2+3+4+.............+31\right).\frac{1}{2}\)
\(\Rightarrow A=495.\frac{1}{2}=\frac{495}{2}\)
Vậy \(A=\frac{495}{2}\)
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