Amount of interest

Question

Find the amount of interest earned between time t and n where $t<n$; if $I_r=r$ for some positive integer r. Answer is $\frac{1}{2}(n^2+n-t^2-t)$

$I_{[t,n]}=A(n)-A(t)$

$I_{[0,r]}=A(r)-A(0)=r$

$A(r)=A(O)+r$

For $t<r<n$

$I_{[r,n]}=A(n)-A(r)$ ;(1)

$I_{[t,r]}=A(r)-A(t)$ ;(2)

Cannot grasp any further concept.

Answer 1

$$ A(n)-A(t)=\sum_{r=t+1}^n I_{[0,r] }=\sum_{r=t+1}^n r = \sum_{r=1}^n r-\sum_{r=1}^t r=\frac{n(n + 1)}{2}-\frac{t(t + 1)}{2}=\frac{1}{2}(n^2 + n- t^2 - t) $$

Answer 2

Let´s assume a simple interest rate of r and an initial amount of 1 monetary unit. And each period the account is increasing by 1 monetary unit.

After the first period you get an interest of $r$. After two periods you get an interest of $r(1+1)=2r$ After $(t-1)$ periods you get an interest of $(t-1)\cdot r$ After n periods you get an interest of $n\cdot r$

The total amount of interest after n periods is $a_n=\color{blue}{r+2r+3r+4r+\ldots+(t-1)r+tr}+\ldots (n-1)r+nr=r(1+2+3+4+\ldots+t+(t+1)+\ldots+(n-1)+n)$

The whole amount (sum) of interest after n periods can be calculated by using the formula for arithmetic series. In this case $a_1=1$ and $a_n=n$.

Therefore $a_n=r\cdot \frac{n\cdot (1+n)}{2} \quad (1)$

Now we have to substract the blue term. Using the formula we have $a_1=1$ and $a_n=t$: $r\cdot \frac{t\cdot (1+t)}{2} \quad (2)$

Subtrating (1) from (2) we get:

$\frac{r}{2} \cdot n\cdot (1+n)-t\cdot (t+1)=\frac{r}{2} \cdot (n^2+n-t^2-t)$

This is not the excact posted solution. But without $r$ it doesn´t make sense.

Answer 3

Answer:

$I_{(0,t)} = A(t) - A(0) $

We also know that:

$I_{(0,1)} = A(1) - A(0) = 1$

$I_{(1,2)} = A(2) - A(1) = 2$

$I_{(2,3)} = A(3) - A(2) = 3$

...

$I_{(t-2,t-1)} = A(t-1) - A(t-2) = t-1$

$I_{(t-1,t)} = A(t) - A(t-1) = t$

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Summing the above, you will see mass cancellation and the sum S reduces to

$S = A(t) - A(0) = I_{(0,t)} = (1+2+3+\cdots|t) = \frac{t(t+1)}{2}\tag{1}$

In a similar reasoning:

$I_{(0,n)} = A(n) - A(0) $

We also know that:

$I_{(0,1)} = A(1) - A(0) = 1$

$I_{(1,2)} = A(2) - A(1) = 2$

$I_{(2,3)} = A(3) - A(2) = 3$

...

$I_{(n-2,n-1)} = A(n-1) - A(n-2) = n-1$

$I_{(n-1,n)} = A(n) - A(n-1) = n$

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Summing the above, you will see mass cancellation and the sum S'reduces to

$S' = A(n) - A(0) = I_{(0,n)} = (1+2+3+\cdots|n) = \frac{n(n+1)}{2}\tag{2}$

Subtract (2) - (1) (i.e S'-S)

We get

$A(n) - A(t) = I_{(t,n)} = \frac{n(n+1)}{2} - \frac{t(t+1)}{2}$

$$ = \frac{1}{2}(n^2+n - t^2 - t)$$