Elements of Algebra

178 ELEMENTS SECT. III. $\frac{105a}{100} = \frac{21a}{20} = \frac{21}{20} \times a, = a + \frac{1}{20}a$. 542. So that, when we add to the original principal its twentieth part, we obtain the amount of the principal at the end of the first year: and adding to this its twentieth part, we know the amount of the given principal at the end of two years, and so on. It is easy, therefore, to compute the successive and annual increases of the principal, and to continue this calculation to any length. 543. Suppose, for example, that a principal, which is at present $1000$l., is put out at five per cent; that the interest is added every year to the principal; and that it were required to find its amount at any time. As this calculation must lead to fractions, we shall employ decimals, but without carrying them farther than the thousandth parts of a pound, since smaller parts do not at present enter into consideration. The given principal of $1000$l. will be worth after 1 year - - - - $1050$l. $52.5,$ after 2 years - - - - $1102.5$ $55.125,$ after 3 years - - - - $1157.625$ $57.881,$ after 4 years - - - - $1215.506$ $60.775,$ after 5 years - - - - $1276.281, \ldots$ which sums are formed by always adding $\frac{1}{20}$ of the preceding principal. 544. We may continue the same method, for any number of years; but when this number is very great, the calculation becomes long and tedious; but it may always be abridged, in the following manner: Let the present principal be $a$, and since a principal of $20$l. amounts to $21$l. at the end of a year, the principal $a$ will amount to $\frac{21}{20} \cdot a$ at the end of a year: and the same principal will amount, the following year, to $\frac{21^2}{20^2} \cdot a = (\frac{21}{20})^2 \cdot a$. Also, this principal of two years will amount to $(\frac{21}{20})^3 \cdot a$, the year after: which will therefore be the principal of three years; and still increasing in the same manner, the given