1/2+1/2.2+1/2.2.2+1/2.2..... 50 thừa số
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a, A = 2 + 2.2 + 2.2.2 + 2.2.2.2 + ... + 2.2...2 ( 22...2 có 16 số 2)
A = 2 + 22 + 23 + 24 + ... + 216
2A = 22 + 23 + 24 + 25 + ... + 217
2A - A = ( 22 + 23 + 24 + 25 + ... + 217) - ( 2 + 22 + 23 + 24 + ... + 216)
A = 217 - 2
b, B = 1 + 1/3 + 1/6 + 1/10 + 1/15 + 1/21 + 1/28
1/2 x B = 1/2 + 1/6 + 1/12 + ... + 1/56
1/2 x B = 1/1x2 + 1/2x3 + 1/3x4 + ... + 1/7x8
1/2 x B = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/7 - 1/8
1/2 x B = 1 - 1/8 = 7/8
B = 7/8 : 1/2
B = 7/8 x 2 = 7/4
a, A = 2 + 2.2 + 2.2.2 + 2.2.2.2 + ... + 2.2...2 ( 22...2 có 16 số 2)
A = 2 + 22 + 23 + 24 + ... + 216
2A = 22 + 23 + 24 + 25 + ... + 217
2A - A = ( 22 + 23 + 24 + 25 + ... + 217) - ( 2 + 22 + 23 + 24 + ... + 216)
A = 217 - 2
b, B = 1 + 1/3 + 1/6 + 1/10 + 1/15 + 1/21 + 1/28
1/2 x B = 1/2 + 1/6 + 1/12 + ... + 1/56
1/2 x B = 1/1x2 + 1/2x3 + 1/3x4 + ... + 1/7x8
1/2 x B = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/7 - 1/8
1/2 x B = 1 - 1/8 = 7/8
B = 7/8 : 1/2
B = 7/8 x 2 = 7/4
đặt tử là T ta có:
2T=2(1+2+22+23+...+22015)
2T=2+22+23+...+22016
2T-T=(2+22+23+...+22016)-(1+2+22+23+...+22015)
T=22016-1
thay T vào tử của S ta được:\(S=\frac{2^{2016}-1}{1-2^{2016}}=-1\)
Đặt A = \(\frac{1}{2}+\frac{1}{2.2}+\frac{1}{2.2.2}+....+\frac{1}{2.2.2.2.2.2.2.2.2.2}\)
=> A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{10}}\)
=> 2A = \(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^9}\)
=> 2A - A = \(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^9}-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
=> A = \(1-\frac{1}{2^{10}}\)
a)
Ta có : \(32^{13}=\left(2^5\right)^{13}=2^{65}\)
\(64^{10}=\left(2^6\right)^{10}=2^{60}\)
Mà \(2^{65}>2^{60}\Rightarrow.....\)
b)
A = 2 + 2.2 + 2.2.2 + ... + 2.2.2.2....2
A = \(2+2^2+2^3+...+2^{100}\)
2A = \(2^2+2^3+2^4+...+2^{101}\)
\(2A-A=\left(2^2+2^3+...+2^{101}\right)-\left(2+2^2+...+2^{100}\right)\)
\(A=2^{101}-2\)
1.
a) Ta có : 3213 = ( 25 ) 13 = 265
6410 = ( 26 ) 10 = 260
Vì 265 > 260 nên 3213 > 6410
b) A = 2 + 2.2 + 2.2.2 + 2.2.2.2 + ... + 2.2.2.2.2...2 ( 100 số 2 )
A = 2 . ( 1 + 2 + 2.2 + 2.2.2 + ... + 2.2.2.2...2 )
A = 2. ( 1 + 2 + 22 + 23 + ... + 299 )
gọi B là biểu thức trong ngoặc
Lại có : B = 1 + 2 + 22 + 23 + ... + 299
2B = 2 + 22 + 23 + 24 + ... + 2100
2B - B = ( 2 + 22 + 23 + 24 + ... + 2100 ) - ( 1 + 2 + 22 + 23 + ... + 299 )
B = 2100 - 1
\(\Rightarrow\)A = 2 . ( 2100 - 1 )
\(\Rightarrow\)A = 2101 - 2
a)10.10.10..10.10.10.10.10.10.10.10.10.10,10,10,10,10.10.10.10.10.10.10.10.10.10.10.10.10 +8=10....08(28 chu so 0).
chia het cho 72 thi phai chia het cho 8va9.
vi 008 chia het cho 8 nen100..8:8
1+0+0+...+0+8=9 chia het cho 9
Vay10.10.....10+8 chia het cho 72 (dpcm)
Đặt A = \(\frac{1}{2}+\frac{1}{2.2}+\frac{1}{2.2.2}+...+\frac{1}{2.2.2.....2}\)
= \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}\)
=> 2A = \(2.\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}\right)\)
= \(2\times\frac{1}{2}+2\times\frac{1}{2^2}+2\times\frac{1}{2^3}+...+2\times\frac{1}{2^{50}}\)
= \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{49}}\)
Lấy 2A - A = \(\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}\right)\)
A = \(1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{50}}\)
= \(1-\frac{1}{2^{50}}\)
Vậy \(\frac{1}{2}+\frac{1}{2.2}+\frac{1}{2.2.2}+...+\frac{1}{2.2.2.....2}\)= \(1-\frac{1}{2^{50}}\)