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Ta có: 2+22+23+24+...+219+220+221
= 2(1+2+22+23+...+220) chia hết cho 2 (1)
Lại có: 2+22+23+24+...+219+220+221
= 2(1+2+22)+...+219(1+2+22)
=2.7+24.7+...+219.7
=7(2+24+...+219) chia hết cho 7 (2)
Mà (2,7)=1 (3)
Từ 1,2,3 => 2+22+23+24+...+219+220+221 chia hết cho 14
Ta có: 2+22+23+24+...+219+220+221
= 2(1+2+22+23+...+220) chia hết cho 2 (1)
có: 2+22+23+24+...+219+220+221
= 2(1+2+22)+...+219(1+2+22)
=2.7+24.7+...+219.7
=7(2+24+...+219) chia hết cho 7 (2)
Mà (2,7)=1 (3)
Từ (1),(2),(3) => 2+22+23+24+...+219+220+221 chia hết cho 14(ddpcm)
Ta có:
\(20A=\frac{20\left(20^{19}+1\right)}{20^{20}+1}=\frac{20^{20}+20}{20^{20}+1}=\frac{20^{20}+1+19}{20^{20}+1}=\frac{20^{20}+1}{20^{20}+1}+\frac{19}{20^{20}+1}=1+\frac{19}{20^{20}+1}\)
\(20B=\frac{20\left(20^{20}+1\right)}{20^{21}+1}=\frac{20^{21}+20}{20^{21}+1}=\frac{20^{21}+1+19}{20^{21}+1}=\frac{20^{21}+1}{20^{21}+1}+\frac{19}{20^{21}+1}=1+\frac{19}{20^{21}+1}\)
Vì 2020+1<2021+1
\(\Rightarrow\frac{19}{20^{20}+1}>\frac{19}{20^{21}+1}\)
\(\Rightarrow1+\frac{19}{20^{20}+1}>1+\frac{19}{20^{21}+1}\)
\(\Rightarrow20A>20B\)
\(\Rightarrow A>B\)
a) Không thể vì: \(\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}=1+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}>1\)
b) Ta có: \(\dfrac{a}{b}< 1\) thì \(\dfrac{a}{b}>\dfrac{a-m}{b-m}\)
CM: \(\dfrac{a}{b}=\dfrac{a\cdot\left(b-m\right)}{b\cdot\left(b-m\right)}=\dfrac{ab-am}{b^2-bm}\left(1\right)\\ \dfrac{a-m}{b-m}=\dfrac{\left(a-m\right)\cdot b}{\left(b-m\right)\cdot b}=\dfrac{ab-am}{b^2-bm}\left(2\right)\)
Vì \(\dfrac{a}{b}< 1\Rightarrow a< b\Rightarrow am< bm\Rightarrow ab-am>ab-bm\left(3\right)\)
Từ (1), (2), (3) ta có \(\dfrac{a}{b}>\dfrac{a-m}{b-m}\)
Vậy
\(B=\dfrac{17^{19}-1}{17^{20}-1}>\dfrac{17^{19}-1-16}{17^{20}-1-16}=\dfrac{17^{19}-17}{17^{20}-17}=\dfrac{17\cdot\left(17^{18}-1\right)}{17\cdot\left(17^{19}-1\right)}=\dfrac{17^{18}-1}{17^{19}-1}=A\)
Vậy B > A
b, B=1+31+....+319
3B=3+32+.....+320
2A=3A-A=320-1
=> A=(320-1):2(đpcm)
2 + 22 + 23 + ... + 219 + 220 + 221
= ( 2 + 22 + 23 ) + ... ( 219 + 220 + 221 )
= ( 2 + 22 + 23 ) + ..... + 219(2 + 22 + 23 )
= 14 + .... + 219.14
= 14 ( 1 + .... + 219) ⋮ 14 (đpcm)
gọi A là tổng
ta có
A = 2 + 22 + 23 + ... + 221
A = ( 2 + 22 + 23 ) + ( 24 + 25 + 26 ) + ... + ( 219 + 220 + 221 )
A = 2 . ( 2 + 22 + 23 ) + 24 . ( 2 + 22 + 23 ) + ... + 219 . ( 2 + 22 + 23 )
A = 2 . 14 + 24 .14 + ... + 219 .14
A = 14 . ( 2 + 24 + ... + 219 ) ( VÌ 14 chia hết cho 14 )
= > A chia hết cho 14