Solve: Consider $f (x) = 90x^2 + 20x + 1$ then sum of digits of $f (111111)$ is...?

Question

This is the question asked in my maths paper of quadratic equations but I am unable to understand which concept will be used here . Please help me in this.

Answer 1

I have calculated the answer. It is 13. $f(1)=111$, sum of digits$=3$.

$f(11)=121*90+20*11+1=11111$, sum of digits$=5$.

In same way when you calculate this sum of digits of

$f(111)=7$

$f(1111)=9$

$f(11111)=11$

$f(111111)=13$

Your answer. The form written is the trick. ($90x^2+20x+1$)

Answer 2

Hint: Complete the square: $$90x^2+20x+1=90(x+\frac19)^2-\frac19=\\ =\frac{10}{9}(9x+1)^2-\frac19=\frac{10(9x+1)^2-1}{9}.$$ Answer: $13$.