giúp mình với !!!!!!!!!!!!!!!!!!!!!!!!

Câu b, bài b1 chứng minh \(a=2^{2006}-1?\)

b.ta chia B thành 10 nhóm mỗi nhóm có 6 hạng tử  \(B=\left(2+2^2+2^3+2^4+2^5+2^6\right)+....+\left(2^{55}+2^{56}+2^{57}+2^{58}+2^{59}+2^{60}\right)\)

\(B\text{=}2\left(1+2+2^2+2^3+2^4+2^5\right)+...+2^{55}\left(1+2+2^2+2^3+2^4+2^5\right)\)

\(B\text{=}2.63+...+2^{56}.63\)

\(\Rightarrow B⋮63\)

\(\Rightarrow B⋮21\)

A=2^1+2^2+2^3+2^4+...+2^2010 

=(2+2^2)+(2^3+2^4)+...+(2^2010+2^2011)

=2.(1+2)+2^3.(1+2)+...+2^2010.(1+2)

=2.3+2^3.3+...+2^2010.3

=(2+2^3+2^2010).3

=> A chia het cho 3

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Mà câu c bạn đánh chia hết thành chết hết rồi kìa

em chịu!!!!!!!!!!!

A= \(\frac{1}{2}\) + \(\frac{1}{2^2}\) + \(\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{100}}\)

\(\Rightarrow\) 2A = 1 + \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}\)

\(\Rightarrow\) 2A - A = ( \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{100}}\) ) -

( \(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}\))

\(\Rightarrow\) A = 1 - \(\frac{1}{2^{100}}\) < 1

Vậy: A < 1
\(\frac{1}{2}\)

B= \(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\)

= 2. \(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

= 2. ( \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\) )

= 2. \(\left(\frac{1}{1}-\frac{1}{100}\right)\) = \(\frac{99}{50}\)

\(\Rightarrow\) B = \(\frac{99}{50}\) < \(\frac{100}{50}\) = 2

Vậy: B < 2

Ta có: \(\frac{1}{2^2}=\frac{1}{4};\frac{1}{3^2}

 ta có \(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)

                ta thấy \(\frac{1}{2^2}