Question

Determine a similar expression for $S_n$. Hence show that $S_n=P{(1+\frac{I}{100})}^n-\frac{100R}{I}({(1+\frac{I}{100})}^n-1)$

Answer 1

$S_1=P(1+\frac{I}{100})-R$ [Here t = 1 & S = P $(1+\frac{x}{100}{)}^t]$

$S_2=P(1+\frac{I}{100}{)}^2-R(1+\frac{I}{100})-R$

$=P(1+\frac{I}{100}{)}^2-R(1+(1+\frac{I}{100}))$

$Sn=P(1+\frac{I}{100}{)}^n-R(1-(1+\frac{I}{100}))+...+(1+\frac{I}{100}{)}^{n-1}$

$=P(1+\frac{I}{100}{)}^n-R\frac{\left(\right(1+\frac{I}{100}{)}^{n-1})}{\frac{I}{100}}$

$P(1+\frac{I}{100}{)}^n-\frac{100R}{I}((1+\frac{I}{100}{)}^n-1)$