Question

\(A=\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{3.5}\right).\left(1+\frac{1}{5.7}\right).....\left(1+\frac{1}{2015.2017}\right)\)

Tính A

 

Answer 1

Lời giải:

Xét tổng quát:

1+1k(k+2)=k(k+2)+1k(k+2)=(k+1)2k(k+2)1+1k(k+2)=k(k+2)+1k(k+2)=(k+1)2k(k+2)

Thay k=1,2,....,2015k=1,2,....,2015 ta có:

1+11.3=221.31+11.3=221.3

1+12.4=322.41+12.4=322.4

1+13.5=423.51+13.5=423.5

1+14.6=524.61+14.6=524.6

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

1+12015.2017=201622015.20171+12015.2017=201622015.2017

Nhân theo vế:

⇒A=12(1+11.3)(1+12.4)(1+13.5)....(1+12015.2017)⇒A=12(1+11.3)(1+12.4)(1+13.5)....(1+12015.2017)

=12.221.3.322.4.423.5.524.6....201622015.2017=12.221.3.322.4.423.5.524.6....201622015.2017

=(1.2.3...2016)2(1.2.3...2015)(2.3.4...2017)=(1.2.3...2016)(2.3....2016)(1.2.3...2015)(2.3.4...2017)=2016.12017=20162017