Deferred Periodicity of a Number

[kaliprasad]

A decimal number is said to be deferred periodicity if it is periodic preceded by one or more digits

show that $\frac{1}{n} + \frac{1}{n+1}+ \frac{1}{n+2}$ forms a decimal fraction of deferred periodicity

 

[MarkFL]

My solution:

Spoiler

At least one of the 3 denominators must be divisible by 2, and so will be a terminating decimal, and at least one of the three denominators must be divisible by 3, and so will be a repeating decimal, and so the sum of the three fractions will be of deferred periodicity regardless of the periodicity of the fraction whose denominator may not be divisible by either 2 or 3.

 

[kaliprasad]

The answer by MARKFL is right

my solution

Spoiler

above sum is

$\dfrac{3n^2+ 6n + 2 }{n(n+1)(n+2)}$

numerator is not divisible by 3 but denominator is divisible

so it is periodic as 3 is not a factor of 10

again if n id odd then numerator odd but denominator is even so 2 divides denominator but not numerator so it is deferred

if n is even $3n^2+6n$ is divisible by 4 but 2 is not so numerator is not divisible by 4 but denominator is divisible by 4 (actually 8) so in reduced for numerator is odd and denominator is even so if is deferred