Question

[|๖ۣۜRPK-ACE(♔⋆şɦเฑเςɦเ⋆♔)]

A=1+3+3^2+3^3+...+3^100

B=1 phần 2+(1 phần 2)^2+(1 phần 2)^3+...+(1 phần 2)^99

C=5^100-5^99+5^98-5^97+...+5^2-5+1

Answer 1

[zZz Cool Kid_new zZz]

a

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

\(3A=3+3^2+3^3+3^4+.....+3^{101}\)

\(2A=3^{101}-1\)

\(A=\frac{3^{101}-1}{2}\)

b

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

\(2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)

\(B=1-\frac{1}{2^{99}}\)

c

\(C=5^{100}-5^{99}+5^{98}-5^{97}+....+5^2-5+1\)

\(5C=5^{101}-5^{100}+5^{99}-5^{98}+....+5^3-5^2+5\)

\(6C=5^{101}+1\)

\(C=\frac{5^{101}+1}{6}\)

Answer 2

[Vũ Cao Minh⁀ᶦᵈᵒᶫ ( ✎﹏IDΣΛ亗 )]

\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}\)

\(\Rightarrow\frac{1}{2}B=\)\(\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{100}\)

\(\Rightarrow B-\frac{1}{2}B=\left[\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}\right]-\left[\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{100}\right]\)

\(\Rightarrow\frac{1}{2}B=\frac{1}{2}-\left(\frac{1}{2}\right)^{100}\Rightarrow B=\left[\frac{1}{2}-\left(\frac{1}{2}\right)^{100}\right].2\)