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a) S = 30 + 32 + 34 + ..... + 32002
9S = 32 + 34 + ..... + 32002 + 32004
9S - S = (32 + 34 + ..... + 32002 + 32004) - (30 + 32 + 34 + ..... + 32002)
8S = 32004 - 30
S = \(\frac{3^{2004}-1}{8}\)
b) S = 30 + 32 + 34 + ..... + 32002
S = (30 + 32 + 34) + (36 + 38 + 310) + ..... + (32000 + 32001 + 32002)
S = (1 + 9 + 81) + 36.(1 + 9 + 81) + ..... + 32000.(1 + 9 + 81)
S = 91 + 36 . 91 + ...... + 32000 . 91
S = 91 . (1 + 36 + ...... + 32000)
S = 7 . 13 . (1 + 36 + ...... + 32000)
Easy????
a) Ta có: S = \(3^0+3^{2^{ }}+...+3^{2002}\)
=> 32S = \(3^2+3^4+3^6+...+3^{2004}\)
=> 9S - S = \(\left(3^2+3^4+3^6+...+3^{2004}\right)-\left(3^0+3^2+...+3^{2002}\right)\)
=> 8S = \(3^{2004}-3^0\)
=> S = \(\dfrac{3^{2004}-1}{8}\)
b) Ta lại có: S = \(3^0+3^{2^{ }}+...+3^{2002}\)
=\(\left(3^0+3^2+3^4\right)+\left(3^6+3^8+3^{10}\right)+....+\left(3^{1998}+3^{2000}+3^{2002}\right)\)
= \(3^0\left(1+3^2+3^4\right)+3^6\left(1+3^2+3^4\right)+....+\)\(3^{1998}\left(1+3^2+3^4\right)\)
= \(91\left(3^0+3^6+...+3^{1998}\right)\)
Vì 91 \(⋮\) 7 => \(91\left(3^0+3^6+...+3^{1998}\right)\) \(⋮\) 7
=> S \(⋮\) 7 ( đpcm)
a, \(S=3^0+3^2+3^4+....+3^{2002}\)
\(3S=3+3^3+....+3^{2003}\)
\(2S=3^{2003}-1\)
b, \(S=\left(3^0+3^2+3^4\right)+\left(3^4+3^6+3^8\right)+...+\left(3^{2000}+3^{1998}+3^{2002}\right)⋮7\)
=> (đpcm)
câu 1hinhf như sai đề
Tớ nghĩ là S= 30 + 32 + 34 +36 +...+ 32002
thì đúng hơn
S=\(3^0+3^2+3^4+...+3^{2002}\)
\(3^2\cdot S=3^2+3^4+3^6+...+3^{2004}\)
9S-S=\(\left(3^2+3^4+3^6+...+3^{2004}\right)-\left(3^0+3^2+3^4+...+3^{2002}\right)\)
8S=\(3^{2004}-3^0\)
8S-\(3^{2004}-1\)=\(3^{2004}-1-3^{2004}-1\)=-2
a)S=30+32+...+32002=1+32+...+32002
=>32.S=32+34+...+32004
=>9S=32+34+...+32004
=>9S-S=(32+34+...+32004)-(1+32+...+32002)
=>8S=32004-1
=>S=\(\frac{3^{2004}-1}{8}\)
b)S=30+32+...+32002=1+32+...+32002
=(1+32+34)+...+(31998+32000+32002)
=91+....+31998.91
=91.(1+...+31998)
=7.13.(1+...+31998) chia hết cho 7
Vậy S chia hết cho 7
\(S=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{11}\left(1+2\right)\)
\(=3\left(2+2^3+...+2^{11}\right)⋮3\)
\(S=2\left(1+2+2^2\right)+...+2^{10}\left(1+2+2^2\right)\)
\(=7\cdot\left(2+...+2^{10}\right)⋮7\)
\(S=3\left(2+2^3+...+2^{11}\right)=3\cdot2\left(1+2^2+...+2^{10}\right)=6\left(1+2^2+...+2^{10}\right)⋮6\)
S = 1 + 2 + 22 + 23 + ... + 220 + 221 (có 22 số; 22 chia hết cho 2)
S = (1 + 2) + (22 + 23) + ... + (220 + 221)
S = 3 + 22.(1 + 2) + ... + 220.(1 + 2)
S = 3 + 22.3 + ... + 220.3
S = 3.(1 + 22 + ... + 220) chia hết cho 3 (đpcm)
\(S=1+2+2^2+2^3+....+2^{21}\)
\(=\left(1+2\right)+2^2\left(1+2\right)+2^4\left(1+2\right)+......+2^{20}\left(1+2\right)\)
\(=\left(1+2\right)\left(1+2^2+2^4+.....+2^{20}\right)\)
\(=3\left(1+2^2+2^4+....+2^{20}\right)\)
Chia hết cho 3
Thái Thùy Dung bn vào câu hỏi tương tự họ giải chi tiết nhá. Nhớ ****. Mk tl sớm nhất royy