Question

Tính :

     \(\frac{2^2}{1.3}+\frac{3^2}{2.4}+\frac{4^2}{3.5}+...+\frac{99^2}{98.100}\)

Help me !!! Ai giúp mik sẽ tik cho...

 

Answer 1

A=2^2/1.3+3^2/2.4+4^2/3.5+....+99^2/98.100

A=2^2/(2-1)(2+1)+3^2/(3-1)(3+1)+4^2/(4-1)(4+1)+...+99^2/(99-1)(99+1)

A=2^2/2^2-1+3^2/3^2-1+...+99^2/99^2-1

A=2^2-1+1/2^2-1+3^2-1+1/3^2-1+...+99^2-1+1/99^2-1

A=1+1/1.3+1+1/2.4+1+1/3.5+...+1+1/98.100

A=(1+1+1+....+1)+(1/1.3+1/2.4+...+1/98.100) (1)

Ta có:

Đặt B=(1+1+1+...+1)=98[vì (99-2):1+1=98 số] (2)

Đặt C=1/1.3+1/2.4+1/3.5+...+1/98.100

=>C=1/2.(1-1/3)+1/2.(1/2-1/4)+1/2.(1/3-1/5)+...+1/2.(1/98-1/100)

=>C=1/2.(1-1/3+1/2-1/4+1/3-1/5+...+1/97-1/99+1/98-1/100)

=>C=1/2.(1+1/2-1/99-1/100)

=>C=1/2.(3/2-1/99.100) (3)

Thay (2),(3) vào(1), được:

A=98+1/2.(3/2-1/99.100)

Answer 2

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