a>

\(\frac{1}{2^2}+\frac{1}{100^2}\)=1/4+1/10000

ta có 1/4<1/2(vì 2 đề bài muốn chứng minh tổng đó nhỏ 1 thì chúng ta phải xét xem có bao nhiêu lũy thừa hoặc sht thì ta sẽ lấy 1 : cho số số hạng )

1/100^2<1/2

=>A<1

oke

A = 1/2 + 1/2^2 + 1/2^3 + ... + 1/2^n < 1

A = 1/2 + 1/2^2 + 1/2^3 + ... + 1/2^n

2A = 1 + 1/2 + 1/2^2+ ..+ 1/2^n-1

2A - A = 1 + 1/2 + 1/2^2+ ..+ 1/2^n-1 - (1/2 + 1/2^2 + 1/2^3 + ... + 1/2^n)

A = (1 - 1/2^n) + (1/2 - 1/2) + ..+ (1/2^n-1 -1/2^n-1)

A = 1 - 1/2^n

A < 1 (đpcm)

Câu b:

B = 1/3 + 1/3^2 + 1/3^3 + ...+ 1/3^n < 1/2

3B = 1 + 1/3 + 1/3^2 + ..+ 1/3^n - 1

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

2B = 1 + 1/3 + 1/3^2 + ..+ 1/3^n - 1 - 1/3 - 1/3^2 - 1/3^3 -..- 1/3^n-1 - 1/3^n

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

2B = 1 - 1/3^n

B = 1/2 - 1/2.3^n < 1/2 (đpcm)

D=\(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)

=>3D=1+\(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)

=>3D-D=(1+\(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\))-(\(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\))

=>2D=1-\(\dfrac{1}{3^{100}}< 1\)

=>D<\(\dfrac{1}{2}\)

Vậy...

D=\(\frac{1}{3}\)+ \(\frac{1}{3^2}\)+\(\frac{1}{3^3}\)+...+\(\frac{1}{3^{100}}\)... Chứng minh D <\(\frac{1}{2}\)

D=\(\frac{1}{1\cdot3}\)+\(\frac{1}{3^2}\)+...+\(\frac{1}{3^{100}}\)< \(\frac{1}{2\cdot3}\)+...+\(\frac{1}{99\cdot100}\)

                                                = \(\frac{1}{2}\)-\(\frac{1}{3}\)+...+\(\frac{1}{99}\)-\(\frac{1}{100}\)

                                                =1/2-\(\frac{1}{100}\)

                                                =49/100 < 1/2

Vậy D <1/2              đpcm

dễ mà bn

Ta có: \(D=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}+\frac{101}{3^{101}}\)

\(\Rightarrow3D=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)

\(\Rightarrow3D-D=\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{101}{3^{101}}\right)\)

\(\Rightarrow2D=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(\Rightarrow6D=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)

\(\Rightarrow6D-2D=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{100}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)

\(\Rightarrow4D=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)

\(\Rightarrow4D< 3-\frac{203}{3^{100}}< 3\Rightarrow D< \frac{3}{4}\left(ĐPCM\right)\)