Question

tìm x, biết

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right):2}=\frac{2013}{2015}\)

Answer 1

2/6+2/12+2/20+...+2/x.(x+1)=2013/2015

2.[1/6+1/12+1/20+...+1/x.(x+1)]=2013/2015

1/2.3+1/3.4+1/4.5+...+1/x.(x+1)=2013/4030

1/2-1/3+1/3-1/4+...+1/x-1/x+1=2013/4030

1/2-1/x+1=2013/4030

1/x+1=1/2015

=> x+1=2015

     x=2014

Vậy x=2014

Answer 2

Đặt A=Vế trái

Ta có :

\(A \over 2 \)\(= \)\({1\over 6 } +{1\over 12 }+{1\over 20 }+...+{1\over x(x+1)}\)

   =\({1\over 2}-{1\over 3}+{1\over 3}-{1\over 4}+{1\over4}-{1\over 5}+...+{1\over x-1}-{1\over x}+{1\over x}-{1\over x+1}\)

   =\({1\over2}-{1\over x+1}\)

Từ đó suy ra: \({1\over2}-{1\over x+1}={2013\over4030}\)

=> x=2014