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\(B=3+3^2+3^3+...+3^{90}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{89}+3^{90}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{89}\left(1+3\right)\)
\(=\left(1+3\right)\left(3+3^3+...+3^{89}\right)\)
\(=4\left(3+3^3+...+3^{89}\right)⋮4\)
\(B=3+3^2+3^3+...+3^{90}\)
\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...\left(3^{88}+3^{89}+3^{90}\right)\)
\(=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{98}\left(1+3+3^2\right)\)
\(=\left(1+3+3^2\right)\left(3+3^4+...+3^{98}\right)\)
\(=13\left(3+3^4+...+3^{98}\right)⋮13\)
\(B=2+2^2+2^3+...+2^{92}\)
=> \(B=(2+2^2+2^3+2^4)+...+\left(2^{89}+2^{90}+2^{91}+2^{92}\right)\)
=> \(B=2(1+2+2^2+2^3)+...+2^{89}\left(1+2+2^2+2^3\right)\)
=> \(B=2.15+...+2^{89}.15\)
=> \(B=(2+...+2^{89}).15\)CHIA HẾT CHO 15
A = 2 + 22 + 23 + ...+ 230
A = ( 2 + 22 ) + ( 23 + 24 ) + ....+ ( 229 + 230 )
A = 2(1+2) + 23(1+2) + ....+ 229(1+2)
A = 2.3 + 23 . 3 + ...+ 229.3
A = 3(2+23 + ...+ 229) \(⋮\) 3
Vậy A chia hết cho 3
1)A=3+32+33+...+32008
A=(3+32)+(33+34)+...+(32007+32008)
A=3(1+3)+33(1+3)+...+32007(1+3)
A=3.4+33.4+...+32007.4
A=4(3+....+32007) chia hết cho 4
\(A=\left(1+2+2^2\right)+\left(2^3+2^4+2^5\right)+...+\left(2^{27}+2^{28}+2^{29}\right)\\ A=\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+...+2^{27}\left(1+2+2^2\right)\\ A=\left(1+2+2^2\right)\left(1+2^3+...+2^{27}\right)\\ A=7\left(1+2^3+...+2^{27}\right)⋮7\)
xem lại đề. số hạng cuối tử số tự nhiên =2; ??? mẫu số cũng ko theo quy luật của 3 số hạng đầu
1, A=(3+3^2)+(3^3+3^4)+...+(3^2007+3^2008)
A= 3.4+3^3.4+...+3^2007 .4
A= 4(3+3^3+...+3^2008)=>ĐPCM
2, theo đề bài :a+b chia hết cho 2
ta có : a+3b=a+b+2b
vì a+b chia hết cho 2 mà 2b chia hết cho 2=> ĐPCM
a, \(A=2+2^2+2^3+...+2^{90}\)
=> \(A=(2+2^2)+(2^3+2^4)+...+(2^{89}+2^{90})\)
=> \(A=2(1+2)+2^3(1+2)+...+2^{89}(1+2)\)
=> \(A=2.3+2^3.3+...+2^{89}.3\)
=> \(A=(2+2^3+...+2^{89}).3\)chia hết cho 3
b, \(A=2+2^2+2^3+...+2^{90}\)
=> \(A=(2+2^2+2^3)+\left(2^4+2^5+2^6\right)+...+(2^{88}+2^{89}+2^{90})\)
=> \(A=2(1+2+2^2)+2^4.\left(1+2+2^2\right)+...+2^{88}(1+2+2^2)\)
=> \(A=2.7+2^4.7+...+2^{88}.7\)
=> \(A=(2+2^4+...+2^{88}).7\)chia hết cho 7
a, A=2+2^2+2^3+2^4+...+2^90
A=(2+2^2)+(2^3+2^4)+..+(2^89+2^90)
A=2.(1+2)+2^3(1+2)+....+2^89(1+2)
A=2.3+2^3.3+...+2^89.3
A=3.(2+2^3+...+2^89)\(⋮\)3
=> A\(⋮\)3=>ĐPCM
b, A=2+2^2+2^3+....+2^90
A=(2+2^2+2^3)+(2^4+2^5+2^6)+...+(2^88+2^89+2^100)
A=2.(1+2+2^2)+2^4.(1+2+2^2)+...+2^88.(1+2+2^2)
A=2.7+2^4.7+...+2^88.7
A=7.(2+2^4+...+2^88)\(⋮\)7
=>A\(⋮\)7=>ĐPCM