Question

已知数列1/1x4,1/4x7,1/7x10……1/(3n-2)(3n+1)

已知数列1/1x4,1/4x7,1/7x10……1/(3n-2)(3n+1), 又因为有一个运算技巧1/a1a2+1/a2a3……1/anan+1=1/a1-1/an+1 , 运用这个技巧解答

Answer 1

原式=1/3*[(1-1/4)+(1/4-1/7)....+(1/(3n-2)-1/（3n+1）)]
=1/3*[1-1/(3n+1)]
=n/(3n+1)

Answer 2

1/(1x4)+1/(4x7)+1/(7x10)+...+1/(3n-2)x(3n+1)]
=1/3*(1-1/4+1/4-1/7+......+1/(3n-2)-1/(3n+1))
=1/3*(1-1/(3n+1))

Answer 3

原式=1/3（1-1/4+1/4-1/7+……+1/（3n-2)-1/(3n+1))
=1/3(1-1/(3n+1))
=n/3n+1

Answer 4

1/1x4+1/4x7+1/7x10+……+1/(3n-2)(3n+1),
=1/3(1-1/4)+1/3(1/4-1/7)+1/3(1/7-1/10)+...+1/3(1/(3n-2)-1/(3n+1))
=1/3((1-1/4+1/4-1/7+1/7-1/10+...+1/(3n-2)-1/(3n+1))
=1/3(1-1/(3n+1))
=1/3(3n/(3n+1))
=n/3n+1