Question

Tính :

S = 4 + 2^{1 ₊} 2² + 2³ + ................... + 2¹⁰⁰

Answer 1

Giải:

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

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

\(\Leftrightarrow2S=2^3+2^2+2^3+2^4+...+2^{101}\)

\(\Leftrightarrow2S-S=\left(2^3+2^{101}\right)-\left(2^2+2^1\right)\)

Hay \(S=2^3+2^{101}-2^2-2^1\)

\(\Leftrightarrow S=8+2^{101}-4-2\)

\(\Leftrightarrow S=2^{101}+2\)

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Answer 2

S = 4 + 2¹ + 2² + 2³ + ... + 2¹⁰⁰

S - 4 = 2¹ + 2² + 2³ + ... + 2¹⁰⁰

2(S - 4) = 2² + 2³ + 2⁴ + 2¹⁰¹

2(S - 4) - (S - 4) = (2² + 2³ + 2⁴ + 2¹⁰¹) - (2¹ + 2² + 2³ + ... + 2¹⁰⁰)

S - 4 = 2¹⁰¹ - 2

S = 2¹⁰¹ + 2