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\(S=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{96}\left(1+3+3^2\right)\)
\(=13+3^3.13+...+3^{96}.13=13\left(1+3^3+...+3^{96}\right)⋮13\)
A=2+22+23+...+299+2100A=2+22+23+...+299+2100
⇒2A=22+23+24+...+2100+2101⇒2A=22+23+24+...+2100+2101
⇒A=2101−2⇒A=2101−2
B=3+32+33+...+399+3100B=3+32+33+...+399+3100
⇒3B=32+33+34+...+3100+3101⇒3B=32+33+34+...+3100+3101
⇒2B=3101−3⇒2B=3101−3
⇒B=3101−32
A=32+33+34+...+397
3A=33+34+35+...+398
3A-A=(33+34+35+...+398)-(32+33+34+...+397)
2A=398-32
A=(398-32): 2
⇒A=(398-32): 2
thế nhé chúc em học tốt :>>☺
ez
+) 32+33+34+...+397
= (32+33)+...+ (396+397)
= 32.(1+3)+...+396.(1+3)
=32.4+...+396.4
=4.(32+...+396)
Vì 4⋮4 nên 4.(32+...+396)⋮4
+)P sau lm như p1 nhx là nhóm 3 số với nhau
Bài 1:
a. $2^{29}< 5^{29}< 5^{39}$
$\Rightarrow A< B$
b.
$B=(3^1+3^2)+(3^3+3^4)+(3^5+3^6)+...+(3^{2009}+3^{2010})$
$=3(1+3)+3^3(1+3)+3^5(1+3)+...+3^{2009}(1+3)$
$=(1+3)(3+3^3+3^5+...+3^{2009})$
$=4(3+3^3+3^5+...+3^{2009})\vdots 4$
Mặt khác:
$B=(3+3^2+3^3)+(3^4+3^5+3^6)+....+(3^{2008}+3^{2009}+3^{2010})$
$=3(1+3+3^2)+3^4(1+3+3^2)+...+3^{2008}(1+3+3^2)$
$=(1+3+3^2)(3+3^4+....+3^{2008})=13(3+3^4+...+3^{2008})\vdots 13$
Bài 1:
c.
$A=1-3+3^2-3^3+3^4-...+3^{98}-3^{99}+3^{100}$
$3A=3-3^2+3^3-3^4+3^5-...+3^{99}-3^{100}+3^{101}$
$\Rightarrow A+3A=3^{101}+1$
$\Rightarrow 4A=3^{101}+1$
$\Rightarrow A=\frac{3^{101}+1}{4}$
`#3107.101107`
\(A = 1 + 3 + 3^2 + 3^3 + ... + 3^{98} + 3^{99}\)
\(A = (1 + 3) + (3^2 + 3^3) + ... + (3^{98} + 3^{99})\)
\(A = (1 + 3) + 3^2(1 + 3) + ... + 3^{98}(1 + 3)\)
\(A = (1 + 3)(1 + 3^2 + ... + 3^{98})\)
\(A = 4(1 + 3^2 + ... + 3^{98})\)
Vì \(4(1 + 3^2 + ... + 3^{98}) \) \(\vdots\) \(4\)
`\Rightarrow A \vdots 4`
Vậy, `A \vdots 4` (đpcm).
A = 1 + 3 + 32 + 33 + ... + 398 + 399
A = (1 + 3) + (32 + 33) + ... + (398 + 399)
A = 1. (1 + 3) + 32. (1 + 3) + ... + 398. (1 + 3)
A = 1.4 + 32.4 + ... + 398.4
A = 4. (1 + 32 + ... + 398)
⇒ A ⋮ 4
cho A = 1 + 3 + 32 + 33 + ... + 311
a ) chứng minh A chia hết cho 13
b) chứng minh A chia hết cho 40
A=1+3+3^2+3^3+...+3^98+3^99+3^100
A=(1+3+ 3^2)+(3^3+3^4+3^5)+...+(3^98+3^99+3^100)
A=(1+3+3^2)+3^3x(1+3+3^2)+...+3^98x(1+3+3^2)
A=13x3^3x13+...+3^98x13
=> 13x(1+3+3^3+...+3^98)chia hết cho 13
Vậy A chia hết cho 13
A = 8⁸ + 2²⁰
= (2³)⁸ + 2²⁰
= 2²⁴ + 2²⁰
= 2²⁰.(2⁴ + 1)
= 2²⁰.17 ⋮ 17
Vậy A ⋮ 17
Ta có:
\(A=1+3+3^2+3^3+3^4+3^5+...+3^{96}+3^{97}+3^{98}\\ \left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{96}\left(1+3+3^2\right)\\ =13+3^3.13+...+3^{96}.13\\ =13\left(1+3^3+3^6+...+3^{96}\right)⋮13\Rightarrow A⋮13\)
Để chứng minh \(A⋮20\) ta chứng minh \(\left\{{}\begin{matrix}A⋮4\\A⋮5\end{matrix}\right.\) vì \(\left(4;5\right)=1\)
Ta có:
+ \(A=1+3+3^2+3^3+3^4+3^5+...+3^{96}+3^{97}+3^{98}\\ =\left(1+3\right)+3^2\left(1+3\right)+3^4\left(1+3\right)+...+3^{96}\left(1+3\right)+3^{98}\\ =4\left(1+3^2+3^4+...+3^{96}\right)+3^{98}\)
\(3^{98}\) có chữ số tận cùng có thể là 1;3;7;9 nên \(3^{98}\) không chia hết cho 4
Vậy A không thể chia hết cho 20.
b.
Ta có:
\(A=1+3+3^2+3^4+3^5+3^6+...+3^{96}+3^{97}+3^{98}\\ \Rightarrow3A=3+3^2+3^4+3^5+3^6+...+3^{96}+3^{97}+3^{99}\\ \Rightarrow3A-A=3^{99}-1\)
\(\Rightarrow A=\dfrac{3^{99}-1}{2}\)
Ta thấy \(3^4=81\Rightarrow3^{99}=3^{96}.3^3=\left(3^4\right)^{24}.3^3=\overline{...7}\)
\(\Rightarrow3^{99}-1=\overline{...7}-1=\overline{...6}\\ \Rightarrow\dfrac{3^{99}-1}{2}=\dfrac{\overline{...6}}{2}=\overline{...3}\)
Đs....