I am trying to prove an identity and quite not get there. The following is the premise.
One deposits $\$1$ at time $t=1,2, \cdots ,n$. evenly spaced. The effective interest per payment is $i$. Let $I_t$ be the interest payable at time $t$. Show that $I_t=(1+i)^{t-1}-1$.
I am thinking that this is an annuity immediate, the accumulated value at time $t$ must be
$$s_{\overline t \rceil i}=\frac{(1+i)^t-1}{i}$$
Similarly that of time $t-1$ is
$$s_{\overline {t-1 }\rceil i}=\frac{(1+i)^{t-1}-1}{i}$$
so since $I_t$ can be simply figured out by
$$s_{\overline {t}\rceil i}-s_{\overline {t-1 }\rceil i} = (1+i)^{t-1}$$
Where is the correct answer getting its $-1$ from?
The only reason I could think of why I am not getting this would be because I am not really understanding the meaning of "interest payable at time $t$". Can someone help me out?
Myszzzzz had the right idea, but I'll outline this with an example.
Suppose in an account of $5\%$ interest I have two payments scheduled to be made: one of $100$ now, and $100$ one year later.
What is the interest from now to one year later? Well, if I put in $100$ now, I get $105$ one year later, so that's interest of $105-100 = 5$.
But, do I include the $100$ one year from now? No. New "principal" payments added to the account are not included.
By that logic, let's go back to this problem.
What's going on at time $1$? I have put in an amount of $1$ into the account.
At time $2$, the $1$ I put in at time $1$ has accumulated value $1(1+i) = 1+i$. BUT, I do not include the principal payment of $1$ that I will be inserting into this account.
Thus, the interest at time $2$ is $I_2 = 1+i-1 = i$.
We can use this information to come up with a more general formula: $$I_t = s_{\overline{t-1}|}(1+i)-s_{\overline{t-1}|} = s_{\overline{t-1}|}(1+i-1) =\dfrac{(1+i)^{t-1}-1}{i}(i)=(1+i)^{t-1}-1\text{.}$$