Thái Thùy Dung bn vào câu hỏi tương tự họ giải chi tiết nhá. Nhớ ****. Mk tl sớm nhất royy

 a, \(S=3^0+3^2+3^4+....+3^{2002}\)

\(3S=3+3^3+....+3^{2003}\)

\(2S=3^{2003}-1\)

b,  \(S=\left(3^0+3^2+3^4\right)+\left(3^4+3^6+3^8\right)+...+\left(3^{2000}+3^{1998}+3^{2002}\right)⋮7\)

=> (đpcm)

Câu 1 đề bài kiểu j thế..bn sửa lại đj

mình đồng ý với lê chí công

S=\(3^0+3^2+3^4+...+3^{2002}\)

\(3^2\cdot S=3^2+3^4+3^6+...+3^{2004}\)

9S-S=\(\left(3^2+3^4+3^6+...+3^{2004}\right)-\left(3^0+3^2+3^4+...+3^{2002}\right)\)

8S=\(3^{2004}-3^0\)

8S-\(3^{2004}-1\)=\(3^{2004}-1-3^{2004}-1\)=-2

a)S=3⁰+3²+...+3²⁰⁰²=1+3²+...+3²⁰⁰²

=>3².S=3²+3⁴+...+3²⁰⁰⁴

=>9S=3²+3⁴+...+3²⁰⁰⁴

=>9S-S=(3²+3⁴+...+3²⁰⁰⁴)-(1+3²+...+3²⁰⁰²)

=>8S=3²⁰⁰⁴-1

=>S=\(\frac{3^{2004}-1}{8}\)

b)S=3⁰+3²+...+3²⁰⁰²=1+3²+...+3²⁰⁰²

=(1+3²+3⁴)+...+(3¹⁹⁹⁸+3²⁰⁰⁰+3²⁰⁰²)

=91+....+3¹⁹⁹⁸.91

=91.(1+...+3¹⁹⁹⁸)

=7.13.(1+...+3¹⁹⁹⁸) chia hết cho 7

Vậy S chia hết cho 7

https://hoc247.net/hoi-dap/toan-6/chung-minh-s-1-2-2-2-2-3-2-4-2-5-2-6-2-7-chia-het-cho-3-faq250754.html

S= \(1+2+2^2+...+2^7\)

2S= \(2\cdot\left(2+2^2+...+2^7\right)\)

2S= \(2^1+2^2+...2^8\)

1S= 2S - S = \(\left(2^1+2^2+...2^8\right)-\left(1+2+2^2+...+2^7\right)\)

1S= \(2^1+2^2+...+2^8-1-2-2^2-...-2^7\)

1S= \(2^8-1\)

1S= \(256-1\)

1S= 255

=> 1S chia hết cho 3

Mà 1S= S

=> S chia hết cho 3

Vậy S chia hết cho 3

Bài 1:

\(A=7+7^3+7^5+...+7^{1999}\)

\(\Rightarrow A=\left(7+7^3\right)+\left(7^5+7^7\right)+...+\left(7^{1997}+7^{1999}\right)\)

\(\Rightarrow A=\left(7+343\right)+7^4\left(7+7^3\right)+...+7^{1996}\left(7+7^3\right)\)

\(\Rightarrow A=350+7^4.350+...+7^{1996}.350\)

\(\Rightarrow A=\left(1+7^4+...+7^{1996}\right).350⋮35\)

\(\Rightarrow A⋮35\left(đpcm\right)\)

b2:

a) \(S=1+3+3^2+...+3^{49}\)

\(\Rightarrow S=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{48}+3^{49}\right)\)

\(\Rightarrow S=\left(1+3\right)+3^2\left(1+3\right)+...+3^{48}\left(1+3\right)\)

\(\Rightarrow S=4+3^2.4+...+3^{48}.4\)

\(\Rightarrow S=\left(1+3^2+...+3^{48}\right).4⋮4\)

\(\Rightarrow S⋮4\left(đpcm\right)\)

c) \(S=1+3+3^2+...+3^{49}\)

\(\Rightarrow3S=3+3^2+3^3+...+3^{50}\)

\(\Rightarrow3S-S=\left(3+3^2+3^3+...+3^{50}\right)-\left(1+3+3^2+...+3^{49}\right)\)

\(\Rightarrow2S=3^{50}-1\)

\(\Rightarrow S=\frac{3^{50}-1}{2}\left(đpcm\right)\)

Giúp mình câu b bài 2 luôn được không?

a)Ta có: \(\frac{3}{1.4}=\frac{4-1}{1.4}=1-\frac{1}{4}\)

\(\frac{3}{4.7}=\frac{7-4}{4.7}=\frac{1}{4}-\frac{1}{7}\)

... . . . .

\(\frac{3}{n\left(n+3\right)}=\frac{1}{n}-\frac{1}{n+3}\)

\(\Leftrightarrow S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+3}< 1^{\left(đpcm\right)}\)

b) Ta có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

   \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)

\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)

Suy ra \(\frac{2}{5}< S\) (1)

Ta lại có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)

Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)

Từ đó suy ra S < 8/9

Từ (1) và (2) suy ra đpcm