Shifted start arithmetic progression formula why it works? $a_n=a_k+(n-k)\cdot d$

Question

Question:The number of zeros in $(10^{60}+1)^2$ is?

The number of zeros in $(10^1+1)^2$ is zero.

The number of zeros in $(10^2+1)^2$ is two.

The number of zeros in $(10^3+1)^2$ is four.

There's a arithmetic progression starting at $n=2$

To figure out the number of zeros I use "$a_n=a_2+(n-2)\cdot 2$"

$a_{60}=2+(60-2)\cdot 2=118$=Answer.

Why does this shifting start arithmetic progression formula I realized work, and what's the generalization?

Answer 1

Does this answer your question?

$a_n=a_1+(n-1) d$

$\ \ \ \ = a_1 +(n-1-k+k)d$

$\ \ \ \ = a_1 + (n-k+k-1)d$

$\ \ \ \ = a_1 + (n-k)d + (k-1)d$

$\ \ \ \ = a_1 +(k-1)d +(n-k)d$

$\ \ \ \ = a_k+(n-k)d \ \ \square$

Edit: It seems I got confused by the wording a bit and this isn't what the OP was asking about at all. I think I'll keep this up for now in case someone has a similar misunderstanding, and also because it'll help anyone that hadn't seen the formula before.

Answer 2

Let's get the number of zeros in $(10^n+1)^2$. Expanding the expression gives us $$10^{2n}+2\cdot 10^n+1.$$

We know that $10^{2n}$ has $2n$ zeros and $2\cdot 10^{n}$ has $n$ zeros. If we add both of them, $10^{2n}+2\cdot 10^{n}$, we will have $2n-1$ zeros because the number $2$ in $2\cdot 10^{n}$ will take the place of one zero at $10^{2n}$. By taking the whole sum, $10^{2n}+2\cdot 10^{n}+1$, we get $2n-2$ because the number $1$ takes the place of the last zero of $10^{2n}+2\cdot 10^{n}$.

So, the number of zeros is $2n-2$, which can be written as $$a_n=2+(n-2)\cdot 2.$$