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B = \(3+3^2+3^3+.....+3^{59}+3^{60}\)
\(=3.\left(1+3\right)+3^3.\left(1+3\right)+....+3^{59}.\left(1+3\right)\)
\(=3.4+3^3.4+....+3^{59}.4\)
\(=4.\left(3+3^3+...+3^{59}\right)⋮4\)
Vậy B chia hết cho 4
Còn phần b) bạn cũng nhóm ra như trên nhưng thêm một số để có tổng là 13
VD : ( 1+3+32)=13 đó
bạn tự làm theo nha
k mik
\(\)
Bạn ơi, sao 23 + 25 mà lại tới 260?
\(1+4+4^2+4^3+...+4^{59}\)
\(=\left(1+4\right)+\left(4^2+4^3\right)+...+\left(4^{58}+4^{59}\right)\)
\(=\left(1+4\right)+4^2.\left(1+4\right)+...+4^{58}.\left(1+4\right)\)
\(=5+4^2.5+...+4^{58}.5\)
\(=5.\left(1+4^2+...+4^{58}\right)⋮5\)
\(\Rightarrow1+4+4^2+4^3+...+4^{59}⋮5\)
\(1+4+4^2+4^3+...+4^{59}\)
\(=\left(1+4+4^2\right)+\left(4^3+4^4+4^5\right)+...+\left(4^{57}+4^{58}+4^{59}\right)\)
\(=\left(1+4+4^2\right)+4^3.\left(1+4+4^2\right)+...+4^{57}.\left(1+4+4^2\right)\)
\(=21+4^3.21+...+4^{57}.21\)
\(=21.\left(1+4^3+...+4^{57}\right)⋮21\)
\(\Rightarrow1+4+4^2+4^3+...+4^{59}⋮21\)
\(1+4+4^2+4^3+...+4^{59}\)
\(=\left(1+4+4^2+4^3\right)+...+\left(4^{56}+4^{57}+4^{58}+4^{59}\right)\)
\(=\left(1+4+4^2+4^3\right)+...+4^{56}.\left(1+4+4^2+4^3\right)\)
\(=85+...+4^{56}.85\)
\(=85.\left(1+...+4^{56}\right)\)
B = (1 + 3) + (32+33)+.....+(389+390)
= 4 + 32 .(1 + 3) + .....+390.(1+3)
= 1 .4 + 32.4 + ..... +390.4
= 4.(1 + 32 + .... +390) chia hết cho 4
\(S=3+3^2+3^3+3^4+....+3^{89}+3^{90}\)
\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{88}+3^{89}+3^{90}\right)\)
\(==3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+3^{88}\left(1+3+3^2\right)\)
\(=\left(1+3+3^2\right).\left(3+3^4+....+3^{88}\right)\)
\(=13\left(3+3^4+...+3^{88}\right)\)\(⋮\)\(13\)
A=1+4+42+...+459A=1+4+42+...+459
A=(1+4)+(42+43)+...+(458+459)A=(1+4)+(42+43)+...+(458+459)
A=(1+4)+42(1+4)+...+458(1+4)A=(1+4)+42(1+4)+...+458(1+4)
A=5+42.5+...+448.5A=5+42.5+...+448.5
A=5(1+42+...+448)A=5(1+42+...+448)
⇒A⋮5
A=1+4+42+...+459A=1+4+42+...+459
A=(1+4+42)+(43+44+45)+...+(457+458+459)A=(1+4+42)+(43+44+45)+...+(457+458+459)
A=(1+4+42)+43(1+4+42)+...+447(1+4+42)A=(1+4+42)+43(1+4+42)+...+447(1+4+42)
A=21+43.21+...+447.21A=21+43.21+...+447.21
A=21(1+43+...+447)A=21(1+43+...+447)
⇒A⋮21
k cho mik đi mik cảm ơn
a) \(\Rightarrow S=\left(1+3\right)+\left(3^2+3^3\right)+.....+\left(3^{88}+3^{99}\right)\)
\(\Rightarrow A=1\left(1+3\right)+3^2\left(1+3\right)+......+3^{88}\left(1+3\right)\)
\(\Rightarrow A=1.4+3^2.4+..........+3^{88}.4\)
\(\Rightarrow A=4.\left(1+3^2+.........+3^{88}\right)\)
Vậy A chia hết cho 4 ĐPCM
b) \(\Rightarrow A=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)\)\(+......+\left(3^{96}+3^{97}+3^{98}+3^{99}\right)\)
\(\Rightarrow A=1\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+\)\(....+3^{96}\left(1+3+3^2+3^3\right)\)
\(\Rightarrow A=1.40+3^4.40+.......+3^{96}.40\)
\(\Rightarrow A=40.\left(1+3^4+....+3^{96}\right)\)
Vậy A chia hết cho 40 ĐPCM
\(B=1+3+3^2+3^3+.....+3^{59}\)
\(B=\left(1+3\right)+3^2\left(1+3\right)+....+3^{58}\left(1+3\right)\)
\(B=4+3^2.4+...+3^{58}.4⋮4\left(đpcm\right)\)
Ta có B = 1 + 3 + 32 + 33 + ... + 359
B = ( 1 + 3 ) + ( 32 + 33 ) + ... + ( 358 + 359 )
B = 1( 1 + 3 ) + 32 ( 1 + 3 ) + ... + 358 ( 1 + 3 )
B = 1 . 4 + 32 . 4 + ... + 358 . 4
B = 4 ( 1 + 32 + ... + 358 ) ⋮ 4 vì 4 ⋮ 4
Vậy B ⋮ 4