tui ko biết tui lớp 4

R.2=2+2^2+2^3+...+2^2015

R=(2+2^2+2^3+...+2^2015-1)-(1+2^2+2^3+...+2^2014)

R=(2^2015)-1

S=(2^2015)-1 / 1-(2^2015)

S=-1

bieu thuc do goi la R nhe

\(S=\left(-2\right)^0+\left(-2\right)^1+\left(-2\right)^2+\left(-2\right)^3+....+\left(-2\right)^{2014}+\left(-2\right)^{2015}\)

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

\(\left(-2\right)S-S=\left[\left(-2\right)+\left(-2\right)^2+...+\left(-2\right)^{2016}\right]-\left[1+\left(-2\right)^1+...+\left(-2\right)^{2015}\right]\)

\(S=\left(-2\right)^{2016}-1\)

( - 2 )²⁰¹⁶ - ( - 2 )⁰

Gọi A = 2^2015 - 2^2014 - ... -2 - 1

2A = 2^2016 - 2^2015 - ... -2^2 - 2

2A - A = 2^2016 - 2 ^2015 - ...-2^2-2 - 2^2015 +2^2014 + .... +2 + 1

A = 2^2016 - 2.2^2015 + 1

A = 2^2016 - 2^2016 + 1

A = 1 

h dung nha

Ta có : 

\(S=2^{2015}-2^{2014}-...-2-1\)

\(S=2^{2015}-\left(2^{2014}+...+2+1\right)\)

Đặt \(A=2^{2014}+...+2+1\) ta có : 

\(2A=2^{2015}+...+2^2+2\)

\(2A-A=\left(2^{2015}+...+2^2+2\right)-\left(2^{2014}+...+2+1\right)\)

\(A=2^{2015}-1\)

\(\Rightarrow\)\(S=2^{2015}-A=2^{2015}-\left(2^{2015}-1\right)=2^{2015}-2^{2015}+1=1\)

Vậy \(S=1\)

Chúc bạn học tốt ~ 

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