Tính nhanh tổng sau:
A = 1 + 2 + 2 ^ 2 + 2 ^ 3 + ... + 2 ^ 1000
B = 3 + 3 ^ 2 + 3 ^ 3 + ... + 3 ^ 2015
C = 4 + 4 ^ 2 + 4 ^ 3 + ... + 4 ^ n
K
Khách
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Những câu hỏi liên quan
29 tháng 11 2023
a: từ 1 đến 100 sẽ có \(\dfrac{100-1}{1}+1=100-1+1=100\left(số\right)\)
=>Sẽ có \(\dfrac{100}{2}=50\) cặp số
1-2+3-4+...+99-100
=(1-2)+(3-4)+...+(99-100)
=(-1)+(-1)+...+(-1)
=-1*50=-50
b: Sửa đề: \(2-4+6-8+...+46-48+50\)
Từ 2 đến 48 sẽ có \(\dfrac{48-2}{2}+1=24-1+1=24\left(số\right)\)
=>Sẽ có \(\dfrac{24}{2}=12\left(cặp\right)\)
\(2-4+6-8+...+46-48+50\)
\(=\left(2-4\right)+\left(6-8\right)+...+\left(46-48\right)+50\)
\(=\left(-2\right)+\left(-2\right)+...+\left(-2\right)+50\)
\(=50-2\cdot24=50-48=2\)
c: Đặt A=\(1+2-3+4+...+97+98-99+100\)
\(=\left(1+2-3+4\right)+\left(5+6-7+8\right)+...+\left(97+98-99+100\right)\)
\(=4+12+...+196\)
Từ 4 đến 196 sẽ có \(\dfrac{196-4}{8}+1=\dfrac{192}{8}+1=25\left(số\right)\)
Tổng của dãy A là: \(\left(196+4\right)\cdot\dfrac{25}{2}=\dfrac{25}{2}\cdot200=100\cdot25=2500\)
H9
HT.Phong (9A5)
CTVHS
15 tháng 8 2023
a) \(1+2+3+4+...+n\)
\(=\left(n+1\right)\left[\left(n-1\right):1+1\right]:2\)
\(=\left(n+1\right)\left(n-1+1\right):2\)
\(=n\left(n+1\right):2\)
\(=\dfrac{n\left(n+1\right)}{2}\)
b) \(2+4+6+..+2n\)
\(=\left(2n+2\right)\left[\left(2n-2\right):2+1\right]:2\)
\(=2\left(n+1\right)\left[2\left(n-1\right):2+1\right]:2\)
\(=\left(n+1\right)\left(n-1+1\right)\)
\(=n\left(n+1\right)\)
c) \(1+3+5+...+\left(2n+1\right)\)
\(=\left[\left(2n+1\right)+1\right]\left\{\left[\left(2n-1\right)-1\right]:2+1\right\}:2\)
\(=\left(2n+1+1\right)\left[\left(2n-1-1\right):2+1\right]:2\)
\(=\left(2n+2\right)\left[\left(2n-2\right):2+1\right]:2\)
\(=2\left(n+1\right)\left[2\left(n-1\right):2+1\right]:2\)
\(=\left(n+1\right)\left(n-1+1\right)\)
\(=n\left(n+1\right)\)
H9
HT.Phong (9A5)
CTVHS
15 tháng 8 2023
d) \(1+4+7+10+...+2005\)
\(=\left(2005+1\right)\left[\left(2005-1\right):3+1\right]:2\)
\(=2006\cdot\left(2004:3+1\right):2\)
\(=2006\cdot\left(668+1\right):2\)
\(=1003\cdot669\)
\(=671007\)
e) \(2+5+8+...+2006\)
\(=\left(2006+2\right)\left[\left(2006-2\right):3+1\right]:2\)
\(=2008\cdot\left(2004:3+1\right):2\)
\(=1004\cdot\left(668+1\right)\)
\(=1004\cdot669\)
\(=671676\)
g) \(1+5+9+...+2001\)
\(=\left(2001+1\right)\left[\left(2001-1\right):4+1\right]:2\)
\(=2002\cdot\left(2000:4+1\right):2\)
\(=1001\cdot\left(500+1\right)\)
\(=1001\cdot501\)
\(=501501\)
NN
3
8 tháng 3 2017
\(A=\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{47\cdot48\cdot49}+\frac{2}{48\cdot49\cdot50}\)
\(A=\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{47\cdot48}-\frac{1}{48\cdot49}+\frac{1}{48\cdot49}-\frac{1}{49\cdot50}\)
\(A=\frac{1}{2\cdot3}-\frac{1}{49\cdot50}\)
\(A=\frac{1}{6}-\frac{1}{2450}\)
\(A=\frac{611}{3675}\)
mong giúp đc bn.thk cho mk
LN
12 tháng 7 2021
\(a,2010:\left(-5\right)+400-1\\ =-402+400-1\\ =-3\\ b,\dfrac{2}{3}+\dfrac{3}{4}.\left(-\dfrac{4}{9}\right)\\ =\dfrac{2}{3}-\dfrac{1}{3}\\ =\dfrac{1}{3}\\ c,\left(1-\dfrac{2}{3}-\dfrac{1}{4}\right)\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\\ =\dfrac{1}{12}.\left(\dfrac{1}{20}\right)^2\\ =\dfrac{1}{12}.\dfrac{1}{400}\\ =\dfrac{1}{4800}\)
12 tháng 7 2021
a) \(2010:\left(-5\right)+400-1=-400+400-1=-1\)
b) \(\dfrac{2}{3}+\dfrac{3}{4}\cdot\dfrac{-4}{9}=\dfrac{2}{3}+\dfrac{-1}{3}=\dfrac{1}{3}\)
c) \(\left(1-\dfrac{2}{3}-\dfrac{1}{4}\right)\cdot\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2=\dfrac{1}{12}\cdot\dfrac{1}{400}=\dfrac{1}{4800}\)
HQ
Hà Quang Minh
Giáo viên
22 tháng 9 2023
a) \(\lim \frac{{2{n^2} + 6n + 1}}{{8{n^2} + 5}} = \lim \frac{{{n^2}\left( {2 + \frac{6}{n} + \frac{1}{{{n^2}}}} \right)}}{{{n^2}\left( {8 + \frac{5}{{{n^2}}}} \right)}} = \lim \frac{{2 + \frac{6}{n} + \frac{1}{n}}}{{8 + \frac{5}{n}}} = \frac{2}{8} = \frac{1}{4}\)
b) \(\lim \frac{{4{n^2} - 3n + 1}}{{ - 3{n^3} + 6{n^2} - 2}} = \lim \frac{{{n^3}\left( {\frac{4}{n} - \frac{3}{{{n^2}}} + \frac{1}{{{n^3}}}} \right)}}{{{n^3}\left( { - 3 + \frac{6}{n} - \frac{2}{{{n^3}}}} \right)}} = \lim \frac{{\frac{4}{n} - \frac{3}{{{n^2}}} + \frac{1}{{{n^3}}}}}{{ - 3 + \frac{6}{n} - \frac{2}{{{n^3}}}}} = \frac{{0 - 0 + 0}}{{ - 3 + 0 - 0}} = 0\).
c) \(\lim \frac{{\sqrt {4{n^2} - n + 3} }}{{8n - 5}} = \lim \frac{{n\sqrt {4 - \frac{1}{n} + \frac{3}{{{n^2}}}} }}{{n\left( {8 - \frac{5}{n}} \right)}} = \frac{{\sqrt {4 - 0 + 0} }}{{8 - 0}} = \frac{2}{8} = \frac{1}{4}\).
d) \(\lim \left( {4 - \frac{{{2^{{\rm{n}} + 1}}}}{{{3^{\rm{n}}}}}} \right) = \lim \left( {4 - 2 \cdot {{\left( {\frac{2}{3}} \right)}^{\rm{n}}}} \right) = 4 - 2.0 = 4\).
e) \(\lim \frac{{{{4.5}^{\rm{n}}} + {2^{{\rm{n}} + 2}}}}{{{{6.5}^{\rm{n}}}}} = \lim \frac{{{{4.5}^{\rm{n}}} + {2^2}{{.2}^{\rm{n}}}}}{{{{6.5}^{\rm{n}}}}} = \lim \frac{{{5^n}.\left[ {4 + 4.{{\left( {\frac{2}{5}} \right)}^{\rm{n}}}} \right]}}{{{{6.5}^n}}} = \lim \frac{{4 + 4.{{\left( {\frac{2}{5}} \right)}^{\rm{n}}}}}{6} = \frac{{4 + 4.0}}{6} = \frac{2}{3}\).
g) \(\lim \frac{{2 + \frac{4}{{{n^3}}}}}{{{6^{\rm{n}}}}} = \lim \left( {2 + \frac{4}{{{{\rm{n}}^3}}}} \right).\lim {\left( {\frac{1}{6}} \right)^{\rm{n}}} = \left( {2 + 0} \right).0 = 0\).
ND
Nguyễn Đức Trí
VIP
5 tháng 8 2023
\(S=\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+\dfrac{1}{4x5}+...\dfrac{1}{nx\left(n+1\right)}\)
\(S=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(S=1-\dfrac{1}{n+1}=\dfrac{n}{n+1}\)
\(T=\dfrac{3}{1x2}+\dfrac{3}{2x3}+\dfrac{3}{3x4}+\dfrac{3}{4x5}+...\dfrac{3}{nx\left(n+1\right)}\)
\(T=3x\left[\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+\dfrac{1}{4x5}+...\dfrac{1}{nx\left(n+1\right)}\right]\)
\(T=3x\left[1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...\dfrac{1}{n}-\dfrac{1}{n+1}\right]\)
\(T=3x\left(1-\dfrac{1}{n+1}\right)=\dfrac{3xn}{n+1}\)
a)A=1+2+22+...+21000
2A=2(1+2+22+...+21000)
2A=2+22+...+21001
2A-A=(2+22+...+21001)-(1+2+22+...+21000)
A=21001-1
b)B=3+32+...+32015
3B=3(3+32+...+32015)
3B=32+33+...+32016
3B-B=(32+33+...+32016)-(3+32+...+32015)
2B=22016-3
\(B=\frac{2^{2016}-3}{2}\)
c)C=4+42+...+4n
4C=4(4+42+...+4n)
4C=42+43+...+4n+1
4C-C=(42+43+...+4n+1)-(4+42+...+4n)
3C=4n+1-4
\(C=\frac{4^{n+1}-4}{3}\)
Ta có: A = 1 + 2 + 22 + ...... + 2100
=> 2A = 2 + 22 + 23 + ...... + 2101
=> 2A - A = 2101 - 1
=> A = 2101 - 1
B = 3 + 32 + 33 + ...... + 22015
=> 3B = 32 + 33 + 34 + ...... + 22016
=> 3B - B = 32016 - 3
=> 2B = 32016 - 3
=> B = 32016 - 3/2
C = 4 + 42 + 43 + .... + 4n
=> 4C = 42 + 43 + 44 + ..... + 4n + 1
=> 4C - C = 4n + 1 - 4
=> 3C = 4n + 1 - 4
=> C = 4n + 1 - 4 / 3