Say someone has total cashflows of $1200$ for the financial year(FY) $2020$, they have cashflows of $1260$ for FY $2021$. ($5\%$ annual growth) How would you go back and fill in the cashflows for Jan-Dec $2021$ to where they are growing at an average annualized rate of $5\%$ and they all total $1260$?
I am assuming steady growth. I've tried pretty basic things, like assuming $\$100$ in cashflows for December $2020$, and then growing that by $100 x 1.05^{\frac{1}{12}}$ ... but this ends up totaling $1232.26$ for $2021$. If I had the December $2020$ cashflow in line with the growth trend, this would be easy. All of the monthly/annual cashflow conversion examples online just say to raise the growth rate by $1/12$.
I've figured out a complicated work around. Going back to a hypothetical year $1$ and assuming $\$100$ for December. Then growing monthly by $1.05^{1/12}$ ... this ends up giving $105$ for Dec year $2$. I can then find the ratio for $105$ over the sum of the year 2 cashflows. $105/1232.25775 = 0.08521...$ I can then use that to get at the Dec $2020$ theoretical cashflow by multiplying it by $1200$... It gives $102.25$... Then if I grow this over the next year, it gives me the correct answer of $1260$... but that seems overly complicated. There has to be a formula for figuring this out right?
Let's assume the months are numbered $0,1,2,\ldots,23$. If your inital month has an amount of $k$ and you want each month to be $r=1.05^{1/12}\approx 1.004074$ times the previous month, then each month you have $kr^n$ and in the first year you get $$k+kr+kr^2+\cdots +kr^{11}= k\frac{r^{12}-1}{r-1} \approx k \times 12.27258$$ so setting this equal to $1200$ gives $k \approx 97.779$, or equivalently in the second year $$kr^{12}+kr^{13}+kr^{14}+\cdots +kr^{23}= k\frac{r^{24}-r^{12}}{r-1} \approx k \times 12.88621$$ so setting this equal to $1260$ again gives $k \approx 97.779$. The final month of your first financial year would be $kr^{11} \approx 102.251$, as you have found. So if you must have a formula for each month, it would be $kr^n = 1200\frac{1.05^{1/12}-1}{1.05-1}1.05^{n/12}$ using months numbered $0,1,2,\ldots,23$