How the formula for EMI is derived

Let $I$ be the installment payment and $B(j)$ the balance remaining after $j$ payments. We want to choose $I$ so that $B(n)=0$. We are given $B(0)=P$. Each month, the interest is applied and the payment deducted to get the new balance, so $B(j)=(1+i)\cdot B(j-1)-I$ If we write this out we get:

$B(0)=P \\ B(1)=(1+i)P-I \\ B(2)=(1+i)((1+i)P-I)-I=(1+i)^2P-(1+i)I-I \\ B(j)=(1+i)^jP-(1+i)^{(j-1)}I-(1+i)^{j-2}I-\ldots I=(1+i)^jP-\frac {(1+i)^j-1}{i}I \\ 0=(1+i)^nP-\frac{(1+i)^n-1}{i}I$

where the second equals sign in $B(j)$ comes from summing the geometric series