Question

Thực hiện phép tính :

a, \(S=2^{2010}-2^{2009}-2^{2008}-...-2-1\)

b, \(P=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{16}\left(1+2+3+...+16\right)\)

Answer 1

\(S=2^{2010}-\left(2^{2009}+2^{2008}+...+2+1\right)\)

Đặt \(A=2^{2009}+2^{2008}+...+2+1\)

\(\Rightarrow2A=2^{2010}+2^{2009}+...+2^2+2\)

\(\Rightarrow2A-2^{2010}+1=2^{2009}+2^{2008}+...+2+1\)

\(\Rightarrow2A-2^{2010}+1=A\)

\(\Rightarrow A=2^{2010}-1\)

\(\Rightarrow S=2^{2010}-A=2^{2010}-\left(2^{2010}-1\right)=1\)

b/ Ta có công thức \(1+2+3+...+n=\dfrac{n\left(n+1\right)}{2}\)

Do đó:

\(P=1+\dfrac{1+2}{2}+\dfrac{1+2+3}{3}+...+\dfrac{1+2+3+...+16}{16}\)

\(P=1+\dfrac{2.3}{2.2}+\dfrac{3.4}{2.3}+\dfrac{4.5}{2.4}+...+\dfrac{16.17}{2.16}\)

\(P=1+\dfrac{1}{2}\left(3+4+5+...+17\right)\)

\(P=1+\dfrac{1}{2}.\dfrac{\left(17-3+1\right)\left(3+17\right)}{2}=76\)