Question

Tính S biết S=\(2+2^2+2^3+...+2^{2016}\)

Answer 1

Ta có : \(2S=2^2+2^3+2^4+...+2^{2017}\)

\(\Rightarrow2S-S=\left(2^2+2^3+2^4+...+2^{2017}\right)-\left(2+2^2+2^3+...+2^{2016}\right)\)

\(\Rightarrow S=\left(2^2+2^3+2^4+...+2^{2016}\right)+2^{2017}-2-\left(2^2+2^3+2^4+...+2^{2016}\right)\)

\(\Rightarrow S=2^{2017}-2\)

Vậy \(S=2^{2017}-2\)

Answer 2

\(S=2+2^2+2^3+...+2^{2016}\)

\(\Rightarrow2S=2^2+2^3+2^4+...+2^{2017}\)

\(\Rightarrow2S-S=\left(2^2+2^3+2^4+...+2^{2017}\right)-\left(2+2^2+...+2^{2016}\right)\)

\(\Rightarrow S=2^{2017}-2\)

Vậy.........

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