a) sửa đề:CM chia 3 dư 1

chia 103 số hạng ra thành nhóm có 3 số

=>103:3=34( nhóm) và dư 1 số hạng

<=> A=\(1+\left(-2+2^2\right.-2^3)+\left(2^4-2^5+2^6\right)-.\ldots+\left(2^{100}-2^{101}+2^{102}\right)\)

A=\(1-2\left(1-2+2^2\right)+2^4\left(1-2+2^2\right)-.\ldots+2^{100}\left(1-2+2^2\right)\)

\(A=1-2\cdot3+2^4\cdot3-.\ldots+2^{100}\cdot3\)

\(A=1+3\left(-2+2^4-.\ldots+2^{100}\right)\)

vì trong ngoặc chia hết cho 3

=> 1 là số còn lại

vậy A chia 3 dư 1

b) \(A=1-2^1+2^2-2^3+\cdots+2^{100}-2^{101}+2^{102}\)

\(2A=2-2^2+2^3-2^4+\cdots+2^{101}-2^{102}+2^{103}\)

=> \(A+2A=\left(1-2^1+2^2-2^3+\cdots+2^{100}-2^{101}+2^{102}\right)+\left(2-2^2+2^3-2^4+\cdots+2^{103}\right)\)

\(3A=1+2^{103}\)

\(A=\frac{\left(2^{103}+1\right)}{3}\)

=> 3D=3¹⁰¹+3¹⁰²+3¹⁰³+...+3¹⁵¹

=> 3D-D=2D=3¹⁵¹-3¹⁰⁰

=> D=\(\frac{3^{151}-3^{100}}{2}\)

\(A=3^2-3^5+3^8-3^{11}+...-3^{101}\)

\(\Rightarrow3A=3^5-3^8+3^{11}-3^{14}+...-3^{104}\)

\(\Rightarrow3A+A=\left(3^5-3^8+3^{11}-3^{14}+...-3^{104}\right)+\left(3^2-3^5+3^8-3^{11}+...-3^{101}\right)\)

\(\Rightarrow4A=-3^{104}+3^2\)

\(\Rightarrow28A=7\left(3^2-3^{104}\right)\)

\(\Rightarrow B+28A=3^{104}+7\left(3^2-3^{104}\right)\)

\(\Rightarrow B+28A=7.3^2-6.3^{104}=3^2\left(7-2.3^{103}\right)\)

Đặt A=1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100

4A=(1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100)4

4A=1.2.3(4-0)+2.3.4(5-1)+3.4.5(6-2)+4.5.6(7-3)+...+98.99.100(101-97)

4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+4.5.6.7-3.4.5.6+...+98.99.100.101-97.98.99.100

4A=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+3.4.5.6-3.4.5.6+...+97.98.99.100-97.98.99.100+98.99.100.101

4A=98.99.100.101

=>A=98.99.100.101/4

=> A=24497550

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