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a) S=(2+22)+22(2+22)+24(2+22)+.....+298(2+22)
S=(2+22)(1+22+24+....+298)
s=6(1+22+24+....+298)
Vi 6 chia het cho 3.Suyra S chia het cho 3
Moi cac ban xem tiep phan sau vao ngay mai
a. S=2+2^2+2^3+2^4+...+2^100
= 2.(1+2)+2^3.(1+2)+2^5.(1+2)+....+2^99(1+2)
=2.3+2^3.3+2^5.3+...+2^99.3
=3.(2+2^2+2^5+...+2^99)
=> 3 chia hết cho 3
b. S=2+2^2+2^3+2^4+...+2^100
= 2.(1+2+4+8)+2^5.(1+2+4+8)+2^9(1+2+4+8)+...+2^96.(1+2+4+8)
=2.15+2^5.15+2^9.15+...+2^96.15
=> S chia hết cho 15
a)Ta có:S = 2^1 + 2^2 + 2^3 + 2^4 + 2^5 +...+2^199+ 2^200.
=( 2^1 + 2^2) + (2^3 + 2^4) + (2^5+2^6)+...+(2^197+2^198)+(2^199+2^200).
=2.(1+2)+2^3.(1+2)+2^5.(1+2)+...+2^197.(1+2)+2^199(1+2)
=2.3+2^3.3+2^5.3+...+2^197.3+2^199.3
=3.(2+2^3+2^5+...+2^197+2^199)
Vậy tổng S chia hết cho 3.
Xin lỗi bn,mik o làm kịp
a) \(S=1+3+3^2+3^3+...+3^{49}\)
\(=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{48}+3^{49}\right)\)
\(=1\left(1+3\right)+3^2\left(1+3\right)+...+3^{48}\left(1+3\right)\)
\(=1.4+3^2.4+...+3^{48}.4\)
\(=\left(3+1\right)\left(1+3^2+...3^{48}\right)=4\left(1+3^2+...+3^{48}\right)⋮4^{\left(đpcm\right)}\)
b) Ta có: \(S=1+3+3^2+3^3+...+3^{49}\)
\(3S=3+3^2+3^3+...+3^{49}+3^{50}\)
\(3S-S=2S=3^{50}-1\Rightarrow S=\frac{3^{50}-1}{2}\)
Ta thấy: \(3^{50}=3^{4.12}.3^2=\left(3^4\right)^{12}.3^2=81^{12}.9=...9\) (tận cùng là 9)
Suy ra \(3^{50}-1=\left(...9\right)-1=...8\) (tận cùng là 8)
Suy ra \(\Rightarrow S=\frac{3^{50}-1}{2}=\frac{\left(...8\right)}{2}=...4\Rightarrow S\) tận cùng là 4
a) \(S=1+3+3^2+3^3+...+3^{49}\)
\(S=\left(1+3\right)+\left(3^2+3^3\right)+....+\left(3^{48}+3^{49}\right)\)
\(S=4+\left(3^2.1+3^2.3\right)+....+\left(3^{48}.1+3^{48}.3\right)\)
\(S=4+3^2.\left(1+3\right)+...+3^{48}.\left(1+3\right)\)
\(S=1.4+3^2.4+...+3^{48}.4\)
\(S=\left(1+3^2+...+3^{48}\right).4⋮4\)
S = \(2+2^2+2^3+...+2^{100}\)
2S = \(2^2+2^3+...+2^{101}\)
2S - S = \(2^{101}-1\)
S = \(2^{101}-1\)
Vì \(101\) chia \(4\) dư \(1\) có dạng \(4k+1\) nên \(2^{101}\)có tận cùng là \(2\) . Mà S = \(2^{101}-1\)nên S có tận cùng là \(1\)
S = \(2+2^2+2^3+...+2^{100}\)
S = \(\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{97}+2^{98}+2^{99}+2^{100}\right)\)
S = \(2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+...+2^{97}\left(1+2+2^2+2^3\right)\)
S = \(3.5.\left(2+2^5+...+2^{97}\right)\)chia hết cho \(3\) và\(5\)