ta có      \(B=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}.\)

      \(\Rightarrow3B=1+\frac{1}{3}+...+\frac{1}{3^{2004}}\)

    \(\Leftrightarrow3B-B=1+\frac{1}{3}-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^2}+...+\frac{1}{3^{2004}}-\frac{1}{3^{2004}}-\frac{1}{3^{2005}}\)

 \(\Leftrightarrow2B=1-\frac{1}{3^{2005}}\)   \(\Rightarrow B=\frac{1-\frac{1}{3^{2005}}}{2}< \frac{1}{2}\left(đpcm\right)\)

Có : 

3B = 1  +1/3 + 1/3^2 + ...... + 1/3^2004

2B = 3B - B = ( 1 + 1/3 + 1/3^2 + ....... + 1/3^2004 ) - ( 1/3 + 1/3^2 + ...... + 1/3^2004 )

     = 1 - 1/3^2004 < 1

=> B < 1/2

Tk mk nha

Bạn tham khảo nhé :) 

Ta có : 

\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\)

\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\)

\(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\right)\) ( loại bỏ các phân số đối nhau ) 

\(2B=1-\frac{1}{3^{2005}}< 1\)

\(B< \frac{1}{2}\)

Vậy \(B< \frac{1}{2}\)

Chúc bạn học tốt ~

Có B=\(\frac{1}{3}\)+\(\frac{1}{3^2}\)+\(\frac{1}{3^3}\)+...+\(\frac{1}{3^{2004}}\)+\(\frac{1}{3^{2005}}\)

=>3B=3.(\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\))

=>3B=1+\(\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)

=>3B-B=(1+\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\))-(\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\))

=>2B=\(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+..+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-....-\frac{1}{3^{2004}}-\frac{1}{3^{2005}}\)

=>2B=1-\(\frac{1}{3^{2005}}\)

=>B=(\(1-\frac{1}{3^{2005}}\)):2

Mà \(\left(1-\frac{1}{3^{2005}}\right)< \frac{1}{2}\)=>\(\left(1-\frac{1}{3^{2005}}\right):2< \frac{1}{2}\)

=>B<\(\frac{1}{2}\)(đpcm)

Sửa đề: Cho \(B=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\). CMR: \(B< \frac{1}{2}\)

Ta có: \(B=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\)

\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\). Lại có:

\(3B-B=2B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\right)\)

\(2B=1-\frac{1}{3^{2005}}< 1\Rightarrow B=\frac{1-\frac{1}{3^{2005}}}{2}< \frac{1}{2}^{\left(đpcm\right)}\)

\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}} \)
\(2B=1-\frac{1}{3^{2004}}\)
\(B=\frac{1}{2}-\frac{1}{2\cdot3^{2004}}\)
Do đó B<\(\frac{1}{2}\)
chúc thành công

1.

A=19^5^1^8^9^0+2^9^1^9^6^9

Ta luôn có 1^a=1 với a là số nguyên dương

=>19^5^1^8^9^0=19⁵ và 2^9^1^9^6^9=2⁹

=>A=19⁵+2⁹=(19²)².19+(2⁴)².2=(...1)².19+(...6)².2=...1.19+...6.2=...1

Vậy A có tận cung là 1.

2.

B=1/3+1/3²+...+1/3²⁰⁰⁵

3B=1+1/3+1/3²+...+1/3²⁰⁰⁴

3B-B=1-1/3²⁰⁰⁵

2B=1-1/3²⁰⁰⁵<1

=>2B<1=>B<1/2

Vậy B<1/2.

.

.

1) Ta có:

\(19^{5^{1^{8^{9^0}}}}+2^{9^{1^{9^{6^9}}}}=19^{5^1}+2^{9^1}\)

Mà 19⁵=19⁴⁺¹=...1.19=...19

      2⁹=2^2.4+1=...6 .2=...2

=>A=...19 + ...2= ...1

Vậy A có chữ số tận cùng là 1

1. 

\(A=19^{5^{1^{8^{9^0}}}}+2^{9^{1^{9^{6^9}}}}\)

Ta có

\(19^{5^{1^{8^{9^0}}}}=0\)

\(2^{9^{1^{9^{6^9}}}}=....6\)

=> \(A=19^{5^{1^{8^{9^0}}}}+2^{9^{1^{9^{6^9}}}}=0+...6=...6\)

=> A có chữ số tận cùng là 6

\(\Rightarrow3B=3+\frac{1}{3^1}+\frac{1}{3^2}+....+\frac{1}{3^{2004}}\)

\(\Rightarrow3B-B=\left(3+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\right)\)

\(\Rightarrow2B=3-\frac{1}{3^{2005}}\Rightarrow B=\left(3-\frac{1}{3^{2005}}\right):2\)

\(\Rightarrow\left(3-\frac{1}{3^{2005}}\right):2<\frac{1}{2}\Rightarrow B<\frac{1}{2}\)

3B=1+1/3+1/3²+...+1/3²⁰⁰⁴

3B-B=1-1/3²⁰⁰⁵

2B=1-1/3²⁰⁰⁵

B=1/2-1/(3²⁰⁰⁵.2)

Vậy B <1/2

Hùng ơi sai rồi

3B=1+1/3+1/3^2+...+1/3^2004 chứ

Thay số 3 thành 1 vì 1/3*3=1 ko phải bằng 3

a) \(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2015}}\)

\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2014}}\)

\(\Rightarrow3B-B=1-\frac{1}{3^{2015}}\)

\(B=\frac{1-\frac{1}{3^{2015}}}{2}\)

giúp câu P luôn với bạn

b) Đặt \(A=\frac{2004}{1}+\frac{2003}{2}+\frac{2002}{3}+...+\frac{1}{2004}\)

\(\Rightarrow A=\left(\frac{2003}{2}+1\right)+\left(\frac{2002}{3}+1\right)+...+\left(\frac{1}{2004}+1\right)+1\) ( tách 2004/1=2004 ra, cộng cho các phân số kia mỗi phân số 1 đơn vị, thì còn dư ra 1)

\(A=\frac{2005}{2}+\frac{2005}{3}+...+\frac{2005}{2004}+\frac{2005}{2005}\) ( 1 = 2005/2005)

\(A=2005.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2004}+\frac{1}{2005}\right)\)

Thay A vào P được

\(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2005}}{2005.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2004}+\frac{1}{2005}\right)}\)

\(P=\frac{1}{2005}\)

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

\(=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)< \sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

\(P=\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+...+\frac{1}{2005\sqrt{2004}}\)

\(\Rightarrow P< 2\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2004}}-\frac{1}{\sqrt{2005}}\right)\)

\(\Rightarrow P< 2\left(1-\frac{1}{\sqrt{2005}}\right)< 2.1=2\)