Question

Tính \(A=1997+P\left(\frac{1}{1998}\right)+P\left(\frac{2}{1998}\right)+...+P\left(\frac{1997}{1998}\right)\)

với \(P\left(x\right)=\frac{2^{2x+1}}{2^{2x}-2}\)và \(x\ne\frac{1}{2}\)

Answer 1

Ta có

\(D=\frac{2^{2x+1}}{2^{2x}-2}+\frac{2^{2\left(1-x\right)+1}}{2^{2\left(1-x\right)}-2}=\frac{2^{2x}}{2^{2x-1}-1}+\frac{2^{2\left(1-x\right)}}{2^{1-2x}-1}\)

Mà \(2^{1-2x}=\frac{1}{2^{2x-1}}\)(do 1-2x+2x-1=0)

=>\(D=\frac{2^{2x}}{2^{2x-1}-1}+\frac{2^{2\left(1-x\right)}}{\frac{1}{2^{2x-1}}-1}=\frac{2^{2x}-2^{2\left(1-x\right)}.2^{2x-1}}{2^{2x-1}-1}=\frac{2^{2x}-2^1}{2^{2x-1}-1}=\frac{2\left(2^{2x-1}-1\right)}{2^{2x-1}-1}=2\)

Áp dụng D ta được

\(P\left(\frac{1}{1998}\right)+P\left(\frac{1997}{1998}\right)=2\)

\(P\left(\frac{2}{1998}\right)+P\left(\frac{1996}{1998}\right)=2\)

..............................................................

Do \(x\ne\frac{1}{2}\)nên không có \(P\left(\frac{999}{1998}\right)\)

\(P\left(\frac{998}{1998}\right)+P\left(\frac{1000}{1998}\right)=2\)

=> \(A=1997+2+2+....+2\)(998 số 2)

=> \(A=1997+2.998=3993\)

Vậy A=3993