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A = 3 + 32 + ... + 3120
= 3(1+3) + 33(1+3) + ... + 3119(1+3)
= 4( 3+ 33 + ... + 3119) chia hết cho 2 (do 4 chia hết cho2)
Vậy ..............................
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a)\(3^{n+2}-2^{n+2}+3^n-2^n=3^n\left(3^2+1\right)-2^n\left(2^2+1\right)\)
\(=3^n.1-2^n.5=3^n.10-2^{n-1}.10=10\left(3^n-2^{n-1}\right)\)nên chia hết cho 10
b)\(9^{120}+9^{119}-9^{118}=9^{118}\left(9^2+9-1\right)=9^{118}.89\)
Suy ra chia hết cho 89
c)\(2^{100}+2^{99}+..+2+1=2^{99}\left(2+1\right)+...+\left(2+1\right)\)
\(=2^{99}.3+2^{97}.3+...+3=3\left(2^{99}+2^{97}+...+1\right)\)nên chia hết cho 3
\(B=3+3^2+3^3+....+3^{120}\)
a, Ta thấy : Cách số hạng của B đều chi hết cho 3
\(B=3+3^2+3^3+....+3^{120}⋮3\)
\(b,B=3+3^2+3^3+....+3^{120}\)
\(B=\left(3+3^2\right)+\left(3^3+3^4\right)+....+\left(3^{119}+3^{120}\right)\)
\(B=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{119}\left(1+3\right)\)
\(B=3.4+3^3.4+...+3^{119}.4\)
\(B=4\left(3+3^3+...+3^{199}\right)\)
Có : \(B=4\left(3+3^3+...+3^{199}\right)⋮4\)
\(\Rightarrow B⋮4\)
\(c,B=3+3^2+3^3+....+3^{120}\)
\(B=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{119}+3^{120}\right)\)
\(B=\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{118}\left(3+3^2\right)\)
\(B=13+3^2.13+...+3^{118}.13\)
\(B=13\left(3^2+3^4+...+3^{118}\right)\)
Có : \(B=13\left(3^2+3^4+...+3^{118}\right)⋮13\)
\(\Rightarrow B⋮13\)
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, mình nghĩ là \(16^5+2^{15}\)
ta có : \(16^5=2^{20}\)
=>\(16^5+2^{15}=2^{20}+2^{15}\)
=\(2^{15}.2^5+2^{15}\)
\(=2^{15}.\left(2^5+1\right)\)
\(=2^{15}.33\)
mà \(2^{15}.33⋮33\)
\(=>16^5+2^{15}⋮33\)