What is the next term of the series? $76, 80, 88, 95, 100, 101, \ldots$ 76, 80, 88, 95, 100, 101, ?

A) 102

B) 103

C) 105

D) 106

The correct answer is given as 103 but I couldn't found out the logic. I have discussed with my friends and they told that the series is wrong.
Can someone confirm if it's wrong or if it's correct then what's the logic?
 A: The logic is quite evident. Consider the polynomial $$f(x) = 103 - \frac{1883 x}{30} + \frac{18551 x^2}{360} - \frac{309 x^3}{16} + \frac{569 x^4}{144} - \frac{101 x^5}{240} + \frac{13 x^6}{720}.$$ You can verify that $f(1) = 76, f(2) = 80,f(3)= 88, \dots ,f(6)= 101$, and.... $$f(7)=103.$$
Now you see it, right? It has to be 103. Why haven't you thought of that formula? Except... this formula: $$102 - \frac{3619 x}{60} + \frac{1971 x^2}{40} - \frac{439 x^3}{24} + \frac{89 x^4}{24} - \frac{47 x^5}{120} + \frac{x^6}{60},$$ that gives 76, 80, 88, ... as usual, but gives 102 as the 7th number! Which should be the correct logic?
OK enough fun. The truth is such patterns are very fuzzy and hard to define, as soon as the pattern becomes too complex to figure out at a glance. You may as well come up with your own version of rules and obtain arbitrary answer.
A: $7+6=13$ and $13=1+3=4$ and $76+4=80$
$8+0=8$ and $80+8=88$
$8+8=16$ and $1+6=7$ and $88+7=95$
$9+5=14$ and $1+4=5$ and $95+5=100$
$1+0+0=1$ and $100+1=101$
$1+0+1=2$ and $101+2=103$
These puzzles are generally not that mathematically clever, always number of digits, or sum of digits, or finite differences, or arithmetic progression.
We have $$x_{n+1}=x_n+\operatorname{sumdigits}^{\infty}(x_n)$$
Where the $\infty$ means sum the digits iteratively until it is stationary.
