Trying to make the problem as simple as possible.
I decide to start on January $1^{\text{st}}$ a saving plan for which the bank will give me a constant interest rate $r$ (the interests being paid on December $31^{\text{st}}$).
During a given year $y$, I shall make a constant monthly deposit $d_y$.
But, this deposit will change from year to year; for example $$d_{y+1}=d_y +\delta\qquad \text{or}\qquad d_{y+1}=d_y \,(1+\delta)\qquad \text{with}\qquad \delta>0$$
What could be the formula of the present value after $n$ years ?
I assume you insert the cash at the end of the year.
The result is $$FV=\frac{d_1}{r-\delta}((1+r)^n-(1+\delta)^n)$$
This is done as follows. We have
$$\forall y\geq 1:d_y=(1+\delta)^{y-1}d_1$$
The cash flow is thus $$d_1,d_2=(1+\delta)d_1,d_3=(1+\delta)^2d_1,\dots$$
So, present value of the cash flow would be $$PV=\sum\limits_{y=1}^n\frac{d_y}{(1+r)^{y}}=\sum\limits_{y=1}^n\frac{d_1(1+\delta)^{y-1}}{(1+r)^{y}}=\frac{d_1}{r-\delta}\Big(1-\frac{(1+\delta)^n}{(1+r)^n}\Big)$$
Using sum of geometric sequence. The future value is $$FV=(1+r)^nPV$$
I further explain. For year $y$ the payment $d_y$ gets $n-y$ years of interest. So, if we only look at what happens with this sum of money we get the following $$d_y,(1+r)d_y,(1+r)^2d_y,\dots,(1+r)^{n-y}d_y$$