A unique Addition Sequnce!

Note that there are ten terms on the RHS, and that $55$ is the sum of the first $10$ natural numbers, not including $0$ . Thus the first term on the RHS must be equal to $1$ the second equal to $2$ , the third equal to $3$ , etc. Solving for $x$ in ten first term, we get $x = 2016$ . Because $x$ is constant throughout the entire problem, we have just solved for x in the entire problem. So, $\boxed{x = 2016}$

I would like to thank TheOnly Caosun, who helped me with this problem.

Great problem, @Satyajit Mohanti

On the LHS there are 10 terms and the RHS is 55, which is the sum of the first 10 natural numbers. Thus we can seperate 55 into 1 , 2 , 3 and so on... to make an equation like this. :

$\frac{x - 2014}{2}$ + $\frac{x-2010}{3}$ + $.....$ $=$ $1 + 2 + .....$

If we subtract 1 on the RHS from the first term on the LHS , we will get:

$\frac{x-2014}{2}$ $- 1$ $=$ $\frac{x - 2014 - 2}{2}$ $=$ $\frac{x - 2016}{2}$

Similarly , if we subtract 2 from the second term on the LHS , we will get:

$\frac{x-2010}{3}$ $- 2$ $=$ $\frac{x - 2010 - 2*3}{3}$ $=$ $\frac{x - 2016}{3}$

Doing this to all terms, we will get this equation:

$\frac{x - 2016}{2}$ $+$ $\frac{x - 2016}{3}$ $+$ $\frac{x - 2016}{4}$ $+$ $............$ $=$ $0$

Since the term $x - 2016$ is constant on the LHS , even taking the LCM we will ultimately obtain this equation:

$x - 2016 = 0$

Thus solving for the equation, we will ultimately obtain:

$\boxed{x = 2016}$