Question

[𝟸𝟿_𝟸𝟷]

Cần giúp '-'

Answer 1

[Giang シ)]

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

\(=>2.E=2+2^2+2^3+...2^{101}\)

\(=>2.M-M=M=201^{101}-1\)

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

\(=>3^2.F=3^2+3^4+3^6+...+3^{102}\)

\(=>F.N-N=3^{102}-1\)

\(=>F=\frac{3^{102}-1}{8}\)

~ Chúc bn hok tốt ~ !!!! 

Answer 2

[Xyz OLM]

e) Ta có : E = 2 + 2² + 2³ + 2⁴ + ... + 2¹⁰⁰

=> 2E = 2² + 2³ + 2⁴ + 2⁵ + ... + 2¹⁰¹

=> 2E - E = (2² + 2³ + 2⁴ + 2⁵ + ... + 2¹⁰¹) - (2 + 2² + 2³ + 2⁴ + ... + 2¹⁰⁰)

=> E = 2²⁰¹ - 2

Vậy E = 2²⁰¹ - 2

f) Ta có F = 1 + 3 + 3² + 3³ + ... + 3¹⁰⁰

=> 3F = 3(1 + 3 + 3² + 3³ + ... + 3¹⁰⁰)  = 3 + 3² + 3³ + 3⁴ + ... + 3¹⁰¹

=> 3F - F = (3 + 3² + 3³ + 3⁴ + ... + 3¹⁰¹) - (1 + 3 + 3² + 3³ + ... + 3¹⁰⁰)

=> 2F = 3¹⁰¹ - 1

=> F = \(\frac{3^{101}-1}{2}\)

Vậy F = \(\frac{3^{101}-1}{2}\)