# Compound interest problem

#### Felwinter

##### New member
HI everybody, and happy new year!

I have a couple questions regarding how to calculate compounded interest. Apologies beforehand because English is not my first language, and my message may be confusing. If so, let me know and I'll try to explain.

1.- I have a $5000 capital (P), which I'm investing for 10 years (t) at a 30% (r) annual (n) interest growth. How can I calculate what will my final capital (A) will be? I know that the formula for compound interest is A = P*(1+(r/n))nt so A =$5000 * (1+(0,3/1))1*10 = $68929,24 2.- I have a$5000 capital (P), to which I'm adding $100 every month, and I'm investing it for 10 years (t) at a 30% (r) annual (t) interest growth. How can I calculate what will my final capital (A) will be? I'm lost here, I can do this manually, but I cannot deduce the formula for this. 3.- I have a$5000 capital (P) that I want to increase by 50% (A = $7500) in 5 years (t) with constant annual (n) return (r). How can I calculate how much (in a percentage) I have to grow it each year to reach that final 50%? Using the previous equation (and n=1), I solve in Wolfram and get R = (3/2)^(1/5) - 1 ≈ 0,0844718 I'm stuck in #2 Thanks! #### firemath ##### Full Member You are only stuck on #2? In that case P+1200 every year Ans x 1.1=Yearly Production. Right? More neatly: (Beginning amount +1200) x 1.1= Year's Production (YP) Next year, you would of course substitute YP for (Beginning amount). Last edited: #### MarkFL ##### Super Moderator Staff member 2.- I have a$5000 capital (P), to which I'm adding \$100 every month, and I'm investing it for 10 years (t) at a 30% (r) annual (t) interest growth. How can I calculate what will my final capital (A) will be?
I would begin by writing the following recursion:

$$\displaystyle A_{n}=\left(1+\frac{r}{12}\right)A_{n-1}+D$$

I am assuming the interest is compounded every month. $$D$$ is the monthly deposit made.

The homogeneous solution is:

$$\displaystyle h_n=c_1\left(1+\frac{r}{12}\right)^n$$

The particular solution will have the form:

$$\displaystyle p_n=c_2$$

Substituting into the recursion, we obtain:

$$\displaystyle c_2=\left(1+\frac{r}{12}\right)c_2+D$$

$$\displaystyle c_2\left(1-\left(1+\frac{r}{12}\right)\right)=D$$

$$\displaystyle c_2=-\frac{12D}{r}$$

And so by the principle of superposition, the closed form will be:

$$\displaystyle A_n=h_n+p_n=c_1\left(1+\frac{r}{12}\right)^n-\frac{12D}{r}$$

We know initially we must have:

$$\displaystyle A_0=c_1-\frac{12D}{r}=P\implies c_1=P+\frac{12D}{r}$$

And so we obtain the formula:

$$\displaystyle A_n=\left(P+\frac{12D}{r}\right)\left(1+\frac{r}{12}\right)^n-\frac{12D}{r}$$

Can you proceed?

#### firemath

##### Full Member
Correction to my original equation: use 1.3 instead of 1.1.

#### firemath

##### Full Member
.....the interest is compounded every month.
Wait.......every month? But it's annual interest. I think that I answered wildly incorrect.

#### MarkFL

##### Super Moderator
Staff member
Wait.......every month? But it's annual interest. I think that I answered wildly incorrect.
I was assuming a monthly compounding of interest. If the interest in compounded annually, then I would say you've given a correct recursion:

$$\displaystyle A_n=1.3A_{n-1}+1200$$

Without further feedback from the OP, no one can really say which assumption is correct here.

#### firemath

##### Full Member
I was assuming a monthly compounding of interest. If the interest in compounded annually, then I would say you've given a correct recursion:

$$\displaystyle A_n=1.3A_{n-1}+1200$$

Without further feedback from the OP, no one can really say which assumption is correct here.
Cool. Thanks!