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b)Ta có:\(A=1+\frac{1}{2.\left(1+2\right)}+\frac{1}{3.\left(1+2+3\right)}+...+\frac{1}{16.\left(1+2+3+...+16\right)}\)
\(=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+...+\frac{1}{16}.\left(1+2+3+...+16\right)\)
\(=1+\frac{1}{2}.3+\frac{1}{3}.6+...+\frac{1}{16}.136\)
\(=1+1,5+2+...+8,5\)
\(=\frac{\left(8,5+1\right).\left[\left(8,5-1\right):0,5+1\right]}{2}=76\)
B=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}<\)
B=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
B=\(1-\frac{1}{8}=\frac{8}{8}-\frac{7}{8}=\frac{1}{8}<2\)
Vậy 1/8<2 hay 1/8<16/8
a, S=1+2+22+23+................+263
\(\Rightarrow\)2S=2+22+23+24+.................+264
\(\Rightarrow\)2S-S=(2+22+23+.................+264) - (1+2+22+...............+263)
\(\Rightarrow\)S=264-1
b,S=1+3+32+.................+320
\(\Rightarrow\)3S=3+32+33+...............+321
\(\Rightarrow\)3S-S=(3+32+33+................+321) - (1+3+32+.................+320)
\(\Rightarrow\)2S=321-1
\(\Rightarrow\)S=\(\frac{3^{21}-1}{2}\)
c,Tương tự:4S=4+42+43+...............+450
\(\Rightarrow\)4S-S=450-1
\(\Rightarrow S=\frac{4^{50}-1}{3}\)
S=1+2^2+2^3+.........+2^63
S=2^0+2^1+2^2+.....+2^63
2S=2x(20+21+22+...+263)
2S=21+22+23+24+......+264
2S-S=(21+22+23+24+..........+264)\(-\)(20+21+22+....+263)
1S=264\(-\)20
S=264\(-\)1
Các câu khác tương tự
câu b nhân S với 3
Câu c nhân S với 4
Cơ số bao nhiêu thì nhân với bấy nhiêu
2, tìm x thuộc Z biết :
a, x^2 -(-3 )^2 =16
x^2-9 = 16
x^2 = 25
=> x = 5
b, x^2 + (-4) ^2 =0
x^2 + 16 = 0
x^2 = -16
=> x= -4
Ta có:
A=2+2^2+2^3+2^4+.....+2^100
=> 2A=2^2+2^3+...+2^101
=> 2A-A=A=(2^2+2^3+...+2^101)-(2+2^2+2^3+2^4.....+2^100)
=> A=2^2+2^3+...+2^101-2-2^2-...-2^100
=> A=2^101-2
B=1+3+3^2+3^2+....+3^2009
=> 3B=3+3^2+3^2+....+3^2010
=> 3B-B=2B=3+3^2+3^2....+3^2010-1-3-3^2-3^2-....-3^2009
=> 2B=3^2010-1
=> B=(3^2010-1)/2
C=1+5+5^2+5^3+...+5^1998
=> 5C=5+5^2+5^3+...+5^1999
=> 5C-C=4C=5+5^2+5^3+...+5^1999-1-5-5^2-5^3-...-5^1998
=> 4C=5^1999-1
=> C=(5^1999-1)/4
D=4+4^2+4^3+...+4^n
=> 4D=4^2+4^3+...+4^n+1
=> 4D-D=3D=4^2+4^3+...+4^n+1 - 4-4^2-4^3-...-4^n
=> 3D=4^n+1 - 4
=> 3D=\(\frac{4^{n+1}-4}{3}\)
Ta có : \(A=2+2^2+2^3+.....+2^{100}\)
\(2A=2+2^2+2^3+.....+2^{101}\)
\(2A-A=2^{101}-2\)
\(A=2^{101}-2\)
a) 3A=1.2.3 + 2.3.3 + 3.4.3 +... + n.(n+1).3
=1.2.(3-0) + 2.3.(4-1) + ... + n.(n+1).[(n+2)-(n-1)]
=[1.2.3+ 2.3.4 + ...+ (n-1).n.(n+1)+ n.(n+1)(n+2)] - [0.1.2+ 1.2.3 +...+(n-1).n.(n+1)]
=n.(n+1).(n+2)
=>S=[n.(n+1).(n+2)] : 3
a, (-2).3 +(-4)-7 .0 +1
= - 6 +(-4) -0 +1
=-10 +1
=-9
c, ( -1) .(-2) + (-3) .(-4) -(-2) .(-3)
=2+ 12-6
=14-6
=8
b, (-1) .(-2) .(-3) .(-4) .(-5) :[ (-3) -(-5) ]
=(-1) .(-2) .(-3) .(-4) .(-5) :2
=2.12.5:2
=(2.5) .12:2
=10 .12:2
=120:2
=60