M=2+2²+...+2²⁰

2M=2²+2³+...+2²¹

=>2M-M=(2²+2³+...+2²¹)-(2+2²+...+2²⁰ )

=>M=2²¹-2=2097150 chia hết cho 5

 M = 2+2²+2³+2⁴+....+2²⁰

M=(2+2²+2³+2⁴)+2⁴x(2+2²+2³+2⁴)+....+2¹⁶x(2+2²+2³+2⁴)

M=30+2⁴x30+....+2¹⁶x30

M=30x(1+2⁴+.....+2¹⁶)

​mà 30 chia hết cho 5

=>30x(1+2⁴+......+2¹⁶​) chia hết cho 5

=>M chia hết cho 5

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

​k mình nha

a) tổng S bằng

(2014+4).671:2=677 039

b)n.(n+2013) để mọi số tự nhiên n mà tổng trên chia hét cho 2 thì n=2n

→2n.(n+2013)\(⋮̸\)2

C)M=2+2²+2³+...+2²⁰

=(2+2²+2³+2⁴)+...+(2¹⁷+2¹⁸+2¹⁹+2²⁰)

=(2+2²+2³+2⁴)+...+(2¹⁶.2+2¹⁶.2²+2¹⁶+2³+2¹⁶.2⁴)

=30.1+...+2¹⁶.(2+2²+2³+2⁴)

=30.1+...+2¹⁶.30

=30(1+2⁵+2⁹+2¹³+2¹⁶)\(⋮\)5

c, M= 2 + 2² + 2³ +........2²⁰

Nhận xét: 2+ 2² + 2³ + 2⁴ = 30; 30 chia hết cho 5

Khi đó: M = ( 2+2² + 2³ + 2⁴ ) + (2⁵ + 2⁶ + 2⁷ + 2⁸)+.....+ (2¹⁷+2¹⁸+2¹⁹+2²⁰)

= ( 2+2² + 2³ + 2⁴ ) + 2⁴. ( 2+2² + 2³ + 2⁴ ) +...........+2¹⁶ .( 2+2² + 2³ + 2⁴ )

= 30+2⁴ .30 + 2⁸. 30 +.........+ 2¹⁶.30

= 30.(2⁴ + 2⁸ +.........+2¹⁶) chia hết cho 5 và 30 chia hết cho 5

Vậy M chia hết cho 5

a, M=1/1.2+1/2.3+...+1/49.50
M=1−1/2+1/2−1/3+...+1/49−1/50
M=1−1/50<1

Vậy M<1

\(a,\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\frac{1}{1}-\frac{1}{50}=\frac{49}{50}< 1\)

\(=>M< 1\)

\(M=2+2^3+2^5+2^7+....+2^{51}\)

\(=\left(2+2^3\right)+\left(2^5+2^7\right)+....+\left(2^{49}+2^{51}\right)\)

\(=10+2^4\left(2+2^3\right)+....+2^{48}\left(2+2^3\right)\)

\(=10+2^4.10+...+2^{48}.10\)

\(=10\left(1+2^4+...+2^{48}\right)\Rightarrow M⋮10\)

\(=2.5.\left(1+2^4+...+2^{48}\right)\Rightarrow M⋮5\)

\(M=2+2^3+2^5+2^7+....+2^{51}.\)

\(M+2^{ }=2+2+2^3+2^5+2^7+.....+2^{51}\)

\(=\left(2+2+2^3\right)+\left(2^5+2^7+2^9\right)+....+\left(2^{47}+2^{49}+2^{51}\right)\)

\(=12+2^4\left(2+2^3+2^5\right)+......+2^{46}\left(2+2^3+2^5\right)\)

\(=12+2^4.42+....+2^{46}.42\)

\(=12+7.3.2\left(2^4+...+2^{46}\right)\)

\(\Rightarrow M=\left[12+7.3.2\left(2^4+.....+2^{46}\right)\right]-2\)

\(=10+7.3.2\left(2^4+....+2^{46}\right)\)

Ta có:  \(7.3.2\left(2^4+...+2^{46}\right)⋮7\)mà 10 không chia hết cho 7

Suy M không chia hết cho 7

mọi người thật là nhẫn tâm

chẳng ai giúp mk

TRỜI ƠI!!! AI MS LÀ BN BÈ THỰC SỰ

Ko cs đứa mô trả lời chứ chi

Loại bn bè vs mấy ng chỉ là giả tạo thôi

a) M =1+3+3²+3³+......+3¹¹⁸+3¹¹⁹
M = ( 1+3+3² ) +...+ ( 3¹¹⁷ + 3¹¹⁸+3¹¹⁹ )
M = 1. ( 1+3+3² ) + ... + 3¹¹⁷ . ( 3¹¹⁷ + 3¹¹⁸+3¹¹⁹ )
M = ( 1+3+3² ) .( 1 + ... + 3¹¹⁷ )
M = 13 . ( 1 + ... + 3¹¹⁷ ) \(⋮\) 13 (đpcm )

b) Ta có:
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
...
\(\dfrac{1}{2009^2}< \dfrac{1}{2008.2009}\)
\(\dfrac{1}{2010^2}< \dfrac{1}{2009.2010}\)

=> \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2009^2}+\dfrac{1}{2010^2}\) < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2008.2009}+\dfrac{1}{2009.2010}\) (1)
Biến đổi vế trái:
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2008.2009}+\dfrac{1}{2009.2010}\)

= \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2008}-\dfrac{1}{2009}+\dfrac{1}{2009}-\dfrac{1}{2010}\)
= \(1-\dfrac{1}{2010}\)
= \(\dfrac{2009}{2010}< 1\) (2)

Từ (1) và (2), suy ra :
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2009^2}+\dfrac{1}{2010^2}\) < 1 hay:
N < 1

\(D=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{10^2}\)

\(\Leftrightarrow D=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{10.10}\)

\(\Leftrightarrow D< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)

\(\Leftrightarrow D< \dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{10-9}{9.10}\)

\(\Leftrightarrow D< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(\Leftrightarrow D< 1-\dfrac{1}{10}\)

\(\Leftrightarrow D< \dfrac{9}{10}< \dfrac{10}{10}=1\)

\(\Leftrightarrow D< 1\left(đpcm\right)\)

Các phần còn lại tương tự như a).

\(M=2+2^2+2^3+...+2^{20}\)

\(\)\(M=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{19}+2^{20}\right)\)

\(M=5+2^2.\left(2+2^2\right)+...+2^{18}.\left(2+2^2\right)\)

\(M=5+2^2.\left(2+2^2\right)+...+2^{18}.\left(2+2^2\right)\)

\(M=5+2^2.5+...+2^{18}.5\)

\(M=5.\left(1+2^2+...+2^{18}\right)\)

\(\text{Do }5⋮5\Rightarrow5.\left(1+2^2+...+2^{18}\right)\)

\(\text{Hay A⋮5}\left(đpcm\right)\)

\(\text{Vậy }A⋮5\)

đpcm và gì vậy ạ ?