a, 5M = 5+1+1/5+1/5^2+.....+1/5^2011

4M=5M-M=(5+1+1/5+1/5^2+.....+1/5^2011)-(1+1/5+1/5^2+.....+1/5^2012)

               = 5-1/5^2012

=> M = (5 - 1/5^2012)/4

Tk mk nha

tự làm đi , cần gì ai chỉ âu

ko biết làm nên nói vậy đây

Ta có :

\(\frac{1}{1^2}<\frac{1}{1.2}\)

\(\frac{1}{3^2}<\frac{1}{2.3}\)

.........

\(\frac{1}{2011^2}<\frac{1}{2010.2011}\)

\(\frac{1}{2012^2}<\frac{1}{2011.2012}\)

\(\Rightarrow A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2011^2}\frac{1}{2012^2}<\frac{1}{1.2}+\frac{1}{2.3}+...\frac{1}{2010.2011}+\frac{1}{2011.2012}=1-\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2010}-\frac{1}{2011}+\frac{1}{2011}-\frac{1}{2012}\)

\(=1-\frac{1}{2012}<1\)

\(\Rightarrow A<1\left(1\right)\)

Lại có A > 0 (2)

Từ (1) & (2) có :

0 < A < 1

\(\Rightarrow\) A  Không phải là số tự nhiên

Ta có:

 1/2^2 < 1/1.2

1/3^2 < 1/2.3

...........

1/2011^2 < 1/2010.2011

1/2012^2 < 1/2011.2012

=>A=1/2^2+1/3^2+...+1/2011^2+1/2012^2<1/1.2+1/2.3+...+1/2010.2011+1/2011.2012=1-1/2+1/2-1/3+...+1/2010-1/2011+1/2011-1/2012     =1-1/2012 < 1

=> A < 1 (1)

Lại có; A>0 (2)

Từ (1) và (2) có:

 0 < A < 1

=> A ko phải là STN

k mih nha

kho the .nhin de bai ma lac het ca mat

giúp tớ với

a)

A=1+4+42+...+459A=1+4+42+...+459

A=(1+4+42)+(43+44+45)+...+(457+458+459)A=(1+4+42)+(43+44+45)+...+(457+458+459)

A=(1+4+42)+43(1+4+42)+...+447(1+4+42)A=(1+4+42)+43(1+4+42)+...+447(1+4+42)

A=21+43.21+...+447.21A=21+43.21+...+447.21

A=21(1+43+...+447)A=21(1+43+...+447)

⇒A⋮21
các số như 43,447,459,458........ là 4 mũ và các số đằng sau là số mũ
câu b cũng làm như vậy nhưng dổi các số và kết quả

A=(3^0+3^1+3^2+3^3)+(3^4+3^5+3^6+3^7)+...+(3^2009+3^2010+3^2011+3^2012)

A=40+3^4*(1+3+3^2+3^3)+...+3^2009*(1+3+3^2+3^3)

A-1=40+80*40+...+3^2009*40

A-1=40*(1+80+..+3^2009)

\(A=\left(3+3^2+3^3+3^4\right)+3^4\left(3+3^2+3^3+3^4\right)+...+3^{2008}\left(3+3^2+3^3+3^4\right)\)

\(=120+3^4.120+...+3^{2008}.120=120\left(1+3^4+...+3^{2008}\right)⋮120\)

\(A=\left(3+3^2+3^3+3^4\right)+...+\left(3^{2009}+3^{2010}+3^{2011}+3^{2012}\right)\)

\(A=\left(3+3^2+3^3+3^4\right)+...+3^{2008}\left(3+3^2+3^3+3^4\right)\)

\(A=\left(3+3^2+3^3+3^4\right)\left(1+3^4+...+3^{2008}\right)\)

\(A=120\left(1+3^4+...+3^{2008}\right)⋮120\)

\(A=3+3^2+3^3+...+3^{2012}\)

\(A=\left(3+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+...+\left(3^{2009}+...+3^{2012}\right)\)

\(A=3\left(1+3+3^2+3^3\right)+3^5\left(1+3+3^2+3^3\right)+...+3^{2009}\left(1+3+3^2+3^3\right)\)

\(A=3.40+3^5.40+...+3^{2009}.40\)

\(A=120+3^4.120+...+3^{2008}.120\)

\(A=120\left(1+3^4+...+3^{2008}\right)⋮120\)

Đọc kĩ đề 1 tí là làm dc ngay:

\(A=\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2012^2}\)

\(A< \dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2011.2012}\)

\(A< \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\)

\(A< \dfrac{1}{2}-\dfrac{1}{2012}< 1\)

Vậy \(A< 1\)

A = \(\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2012^2}\)
Ta có :
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
...
\(\dfrac{1}{2012^2}< \dfrac{1}{2011.2012}\)
=> A = \(\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2012^2}\)< \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2011.2012}\) (1)
Biến đổi vế trái :
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2011.2012}\)
= \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\)
= \(\dfrac{1}{2}-\dfrac{1}{2012}\)
= \(\dfrac{1005}{2012}\)< 1 (2)
Từ (1) và (2), suy ra:
A < 1

\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{2012}{3^{2012}}\)

\(\Rightarrow3A=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{2012}{3^{2011}}\)

\(\Rightarrow3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{2012}{3^{2011}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{2012}{3^{2012}}\right)\)

\(\Rightarrow2A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{2011}}-\frac{2012}{3^{2012}}\)

\(\Rightarrow6A=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2010}}-\frac{2012}{3^{2011}}\)

\(\Rightarrow6A-2A=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2010}}-\frac{2012}{3^{2011}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{2011}}-\frac{2012}{3^{2012}}\right)\)

\(\Rightarrow4A=3-\frac{2012}{3^{2011}}\)

\(\Rightarrow A=\frac{3-\frac{2012}{3^{2011}}}{4}=\frac{3}{4}-\frac{\frac{2012}{3^{2011}}}{4}=\frac{3}{4}-\frac{2012}{3^{2011}.4}\)

\(\Rightarrow A< \frac{3}{4}\)

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