\(\frac{-1}{2}\cdot17,5-\frac{2015}{2016}\cdot2018+\frac{1}{2}\cdot7,5+\frac{2015}{2016}\cdot2\)

\(=-\frac{1}{2}\left(17,5-7,5\right)-\frac{2015}{2016}\left(2018-2\right)\)

\(=-\frac{1}{2}\cdot10-\frac{2015}{2016}\cdot2016=-5-2015=-2020\)

Đặt A= 1+2+2²+2³+...+2²⁰¹⁵

<=> 2A=2+2²+2³+...+2²⁰¹⁶

2A-A=(2+2²+2³+...+2²⁰¹⁶)-(1+2+2²+...+2²⁰¹⁵)

<=> A=2²⁰¹⁶-1

B =\(\frac{1+2+2^2+2^3+...+2^{2015}}{1-2^{2016}}\)

=\(\frac{2^{2016}-1}{1-2^{2016}}\)

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

=-1

Vậy B=-1

Phạm Công Thành giúp mik câu hỏi của mik vs !!!!!!!!

\(P=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+....+\frac{1}{2016}.\left(1+2+3+...+2016\right)\)

\(P=1+\frac{1}{2}.3+\frac{1}{3}.6+\frac{1}{4}.10+....+\frac{1}{2016}.2033136\)

\(P=1+\frac{3}{2}+4+\frac{5}{2}+....+\frac{2017}{2}\)

\(P=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+....+\frac{2017}{2}\)

\(P=\frac{2+3+4+5+....+2017}{2}=\frac{2035152}{2}=1017576\)

Ta có : \(\frac{n-1}{n!}=\frac{1}{\left(n-1\right)!}-\frac{1}{n!}\) với n là số tự nhiên khác 0

Khi đó : \(A=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{2015}{2016!}\)

\(=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+...+\frac{1}{2015!}-\frac{1}{2016!}\)

\(=1-\frac{1}{2016!}< 1\)

Lại có B > 1

=> A < B