EXAMPLES

EXAMPLES. 1. Let $\backslash frac\{1\}\{a-x\}$ be converted into an infinite series? Assume $\backslash frac\{1\}\{a-x\}\; =\; A\; +\; Bx\; +\; Cx^2\; +\; Dx^3\; +\; Ex^4$, &c.

Then this assumed series, multiplied by $a-x$, gives $1\; =\; aA\; +\; aBx\; +\; aCx^2\; +\; aDx^3\; +\; aEx^4$, &c.
$-Ax\; -\; Bx^2\; -\; Cx^3\; -\; Dx^4$, &c.

Now, by equating the coefficients of the same powers of $x$, we have $aA\; =\; 1$, $aB\; -\; A\; =\; 0$, $aC\; -\; B\; =\; 0$, $aD\; -\; C\; =\; 0$, $aE\; -\; D\; =\; 0$, &c. Hence $A\; =\; \backslash frac\{1\}\{a\}$, $B\; =\; \backslash frac\{A\}\{a\}\; =\; \backslash frac\{1\}\{a^2\}$, $C\; =\; \backslash frac\{B\}\{a\}\; =\; \backslash frac\{1\}\{a^3\}$, $D\; =\; \backslash frac\{C\}\{a\}\; =\; \backslash frac\{1\}\{a^4\}$, $E\; =\; \backslash frac\{D\}\{a\}\; =\; \backslash frac\{1\}\{a^5\}$, &c.; whence, by substitution, we have $\backslash frac\{1\}\{a-x\}\; =\; \backslash frac\{1\}\{a\}\; +\; \backslash frac\{x\}\{a^2\}\; +\; \backslash frac\{x^2\}\{a^3\}\; +\; \backslash frac\{x^3\}\{a^4\}\; +\; \backslash frac\{x^4\}\{a^5\}$, &c.

2. Convert the quantity $\backslash frac\{c^2\}\{c^2+2cy+y^2\}$ into an infinite series?

Let the assumed series be $A\; +\; By\; +\; Cy^2\; +\; Dy^3$, &c. which multiplied by $c^2\; +\; 2cy\; +\; y^2$, gives $c^2\; =\; c^2A\; +\; c^2By\; +\; c^2Cy^2\; +\; c^2Dy^3$, &c.
$+\; 2cAy\; +\; 2cBy^2\; +\; 2cCy^3$
$-\; Ay^4\; -\; By^5$.

Now, by equating the coefficients of the homologous terms, we have $c^2\; =\; c^2A$, $c^2B\; +\; 2cA\; =\; 0$, $c^2C\; +\; 2cB\; =\; 0$, $c^2D\; +\; 2cC\; =\; 0$, &c.; whence $A\; =\; 1$, $B\; =\; -\backslash frac\{2A\}\{c\}\; =\; -\backslash frac\{2\}\{c\}$, $C\; =\; \backslash frac\{A-2B\}\{c^2\}\; =\; \backslash frac\{1+4\}\{c^2\}\; =\; \backslash frac\{5\}\{c^2\}$, $D\; =\; \backslash frac\{B-2C\}\{c^2\}\; =\; \backslash frac\{-2-10\}\{c^2\}\; =\; -\backslash frac\{12\}\{c^2\}$, &c.; whence $\backslash frac\{c^2\}\{c^2+2cy+y^2\}\; =\; 1\; -\; \backslash frac\{2y\}\{c\}\; +\; \backslash frac\{5y^2\}\{c^2\}\; -\; \backslash frac\{12y^3\}\{c^3\}$, &c.

3. Required the square root of $a^2-x^2$? Let $a^2-x^2\; =\; A\; +\; Bx^2\; +\; Cx^4\; +\; Dx^6$, &c. which being squared gives $a^2\; -\; x^2\; =\; A^2\; +\; 2ABx^2\; +\; B^2x^4\; +\; 2ADx^6$, &c.
$+\; 2ACx^4\; +\; 2BCx^6$.

Hence $A^2\; =\; a^2$, $2AB\; +\; 1\; =\; 0$, $B^2\; +\; 2AC\; =\; 0$, $2AD\; +\; 2BC\; =\; 0$, &c. Then $A\; =\; a$, $B\; =\; -\backslash frac\{1\}\{2A\}\; =\; -\backslash frac\{1\}\{2a\}$, $C\; =\; -\backslash frac\{B^2\}\{2A\}\; =\; \backslash frac\{1\}\{8a^3\}$, $D\; =\; -\backslash frac\{BC\}\{A\}\; =\; \backslash frac\{1\}\{16a^5\}$, &c.; whence $a^2\; -\; x^2\; =\; a^2\; -\; \backslash frac\{x^2\}\{2a\}\; -\; \backslash frac\{x^4\}\{8a^3\}\; -\; \backslash frac\{x^6\}\{16a^5\}$, &c.

As this method has already been illustrated in the article ALGEBRA, we shall therefore briefly state the theorem, and add a few examples.

1. Let $\backslash frac\{a\}\{ax-x^2\}$ be converted into an infinite series? Now $\backslash frac\{a\}\{ax-x^2\}\; =\; a\; \backslash times\; \backslash frac\{1\}\{ax-x^2\}\; =\; \backslash frac\{a\}\{ax\}\; \backslash times\; \backslash frac\{1\}\{1-\backslash frac\{x\}\{a\}\}$. And this last expression, being compared with the general theorem, gives $\backslash frac\{b\}\{a\}\; =\; \backslash frac\{x\}\{a\}$, $m\; =\; -1$, $n\; =\; 2$. Hence, by substitution, we have $\backslash frac\{a\}\{ax-x^2\}\; =\; \backslash frac\{a\}\{ax\}\; \backslash times\; \backslash frac\{1\}\{1-\backslash frac\{x\}\{a\}\}\; =\; \backslash frac\{1\}\{x\}\; \backslash times\; \backslash frac\{1\}\{1-\backslash frac\{x\}\{a\}\}\; =\; \backslash frac\{1\}\{x\}\; \backslash times\; \backslash frac\{1\}\{1-\backslash frac\{x\}\{a\}\}\; =\; \backslash frac\{1\}\{x\}\; \backslash times\; \backslash frac\{1\}\{1-\backslash frac\{x\}\{a\}\}\; =\; \backslash frac\{1\}\{x\}\; \backslash times\; \backslash frac\{1\}\{1-\backslash frac\{x\}\{a\}\}$

2. Required the square root of $a^2\; +\; x^2$?

By comparing this with the general theorem, we have $a=a^2$, $b=x^2$, $m=1$, $n=2$. Hence, by substitution, the series becomes $a\; \backslash times\; 1\; +\; \backslash frac\{1\}\{2\}\; \backslash times\; \backslash frac\{x^2\}\{a^2\}\; +\; \backslash frac\{1-2\}\{2\; \backslash times\; 2\}\; \backslash times\; \backslash frac\{x^4\}\{a^4\}$

In order to apply this to numbers, let the square root of 85 be required? Now, the square root of 85 $=\; \backslash sqrt\{81\; +\; 4\}$; hence $a=9$, and $x^2=4$.

Square root of 85 $=\; 9.219546$, true except the last decimal.

3. Required the cube root of $a^3\; +\; b^3$?

This being compared with the general theorem gives $a=a^3$, $b=b^3$, $m=1$, $n=3$. Hence $a^2\; +\; b^2\; \backslash frac\{1\}\{2\}\; =$

Let the cube root of 600 be required? Now $600\; \backslash frac\{1\}\{2\}\; =\; 8\; \backslash times\; 1\; +\; \backslash frac\{1\}\{2\}\; \backslash frac\{1\}\{12\}\; \backslash frac\{1\}\{2\}$. Then $y^3=88$, $x^3=512$, $m=1$, and $n=3$.

Cube root of 600 $=\; 8.43432664$

In operations of this kind, the nearest power to the given number, whether greater or less than it, is to be used, as by that means the series will converge more quickly.

An infinite series may be involved to any given involution power, or any proposed root of a given series may be extracted by means of the following general theorem.

$z^m \times (a + b x + c x^2 + d x^3 + e x^4 + f x^5 + g x^6 + h x^7 + i x^8 + j x^9 + k x^{10} + l x^{11} + m x^{12} + n x^{13} + o x^{14} + p x^{15} + q x^{16} + r x^{17} + s x^{18} + t x^{19} + u x^{20} + v x^{21} + w x^{22} + x x^{23} + y x^{24} + z x^{25} + a x^{26} + b x^{27} + c x^{28} + d x^{29} + e x^{30} + f x^{31} + g x^{32} + h x^{33} + i x^{34} + j x^{35} + k x^{36} + l x^{37} + m x^{38} + n x^{39} + o x^{40} + p x^{41} + q x^{42} + r x^{43} + s x^{44} + t x^{45} + u x^{46} + v x^{47} + w x^{48} + x x^{49} + y x^{50} + z x^{51} + a x^{52} + b x^{53} + c x^{54} + d x^{55} + e x^{56} + f x^{57} + g x^{58} + h x^{59} + i x^{60} + j x^{61} + k x^{62} + l x^{63} + m x^{64} + n x^{65} + o x^{66} + p x^{67} + q x^{68} + r x^{69} + s x^{70} + t x^{71} + u x^{72} + v x^{73} + w x^{74} + x x^{75} + y x^{76} + z x^{77} + a x^{78} + b x^{79} + c x^{80} + d x^{81} + e x^{82} + f x^{83} + g x^{84} + h x^{85} + i x^{86} + j x^{87} + k x^{88} + l x^{89} + m x^{90} + n x^{91} + o x^{92} + p x^{93} + q x^{94} + r x^{95} + s x^{96} + t x^{97} + u x^{98} + v x^{99} + w x^{100}) = z^m$, multiplied by

Now each term of the given series is to be compared with the correspondent terms in the first part of the

above theorem; and by substitution in the second, the several terms of the required series will be obtained.

1^st. What is the square of the series $y\; -\; y^2\; +\; y^3\; -\; y^4\; +\; \backslash dots$?

By comparing this with the general theorem, we find $z\; =\; y$, $a\; =\; 1$, $b\; =\; 0$, $c\; =\; -1$, $d\; =\; 0$, $g\; =\; -1$, &c. and $m\; =\; 2$; whence $y\; -\; y^2\; +\; y^3\; -\; y^4\; +\; \backslash dots\; =\; y^2\; \backslash times\; (1\; -\; 2ax^2\; +\; c^2x^4\; -\; 2ce^2x^6\; +\; 2ex^8\; -\; 2gx^\{10\}\; +\; \backslash dots)$, &c. $=\; y^2\; \backslash times\; (1\; -\; 2y^2\; +\; 3y^4\; -\; 4y^6\; +\; \backslash dots)$, &c. $=\; y^2\; -\; 2y^4\; +\; 3y^6\; -\; 4y^8\; +\; \backslash dots$.

2^d. Required the fourth power of the series $1\; +\; x\; +\; x^2\; +\; x^3\; +\; \backslash dots$?

Here $z\; =\; 1$, $a\; =\; 1$, $b\; =\; 1$, $c\; =\; 1$, $d\; =\; 1$, & $m\; =\; 4$.

3^d. What is the square of $\backslash frac\{1\}\{x\}\; +\; \backslash frac\{1\}\{x^2\}\; +\; \backslash frac\{1\}\{x^3\}\; +\; \backslash frac\{1\}\{x^4\}\; +\; \backslash dots$?

In this case $z\; =\; \backslash frac\{1\}\{x\}$, $x\; =\; \backslash frac\{1\}\{x\}$, $a\; =\; 1$, $b\; =\; 1$, $c\; =\; 1$, $d\; =\; 1$, & $m\; =\; 2$.

4^th. What is the square root of $\backslash frac\{1\}\{r^2\; -\; \backslash frac\{z^2\}\{2\}\; +\; \backslash frac\{z^4\}\{4r^2\}\; -\; \backslash frac\{z^6\}\{6r^4\}\; +\; \backslash frac\{z^8\}\{8r^6\}\; +\; \backslash dots\}$?

The quantity reduced is $\backslash frac\{1\}\{r^2\}\; \backslash times\; \backslash frac\{1\}\{1\; -\; \backslash frac\{z^2\}\{2r^2\}\; +\; \backslash frac\{z^4\}\{4r^2\}\; -\; \backslash frac\{z^6\}\{6r^4\}\; +\; \backslash frac\{z^8\}\{8r^6\}\; +\; \backslash dots\}$.

In this example $z\; =\; \backslash frac\{1\}\{r^2\}$, $x\; =\; \backslash frac\{1\}\{r^2\}$, $a\; =\; 1$, $b\; =\; -\backslash frac\{1\}\{2r^2\}$, $c\; =\; \backslash frac\{1\}\{4r^2\}$, $d\; =\; -\backslash frac\{1\}\{6r^2\}$, &c. and $m\; =\; -\backslash frac\{1\}\{2\}$, $\backslash frac\{m-1\}\{2\}\; =\; -\backslash frac\{3\}\{4\}$, $\backslash frac\{m-2\}\{3\}\; =\; -\backslash frac\{5\}\{6\}$, $\backslash frac\{m-3\}\{4\}\; =\; -\backslash frac\{7\}\{8\}$, &c.

Of an harmonic series, a series of terms formed in harmonicical proportion. It has been already observed in the article PROPORTION, that if three numbers be in harmonicical proportion, the first is to the third as the difference between the first and second is to the difference between the second and third.

Let $a$, $b$, and $x$ be three terms in harmonicical proportion: then $a\; :\; x\; :\; b\; =\; a\; -\; b\; :\; b\; -\; x$, whence $a\; \backslash times\; b\; -\; b\; \backslash times\; a\; =\; a\; \backslash times\; x$, and $2ax\; -\; bx\; =\; ab$.

then $x\; =\; \backslash frac\{ab\}\{2a-b\}$. Hence the three

first terms of this series is $a,\; b,\; \backslash frac\{ab\}\{2a-b\}$.

Again, let $x$ be the fourth term, to find which in terms of $a$ and $b$, we have

therefore the four first terms are $a,\; b,\; \backslash frac\{ab\}\{2a-b\},\; \backslash frac\{ab\}\{3a-2b\}$.

Whence the law of the series is obvious, and it may be continued

Series continued as follows, and the $n^\{\backslash text\{th\}\}$ term is $\backslash frac\{ab\}\{n-1.a-n-2.b\}$

If, in a series of terms in harmonical proportion, $a$ and $b$ be two affirmative quantities, and such that $a\; \backslash leq\; b$; then this series, which is positive at first, will become negative as soon as $n-2.b$ exceeds $n-1.a$. But if $a\; >\; b$, the series will converge, and although produced to infinity will not become negative.

Let $a$ and $b$ be equal to 2 and 1 respectively; then this series becomes and since, if each term of an harmonical series be divided by the same quantity, the series will still be harmonical. Therefore is an harmonical series: whence the denominators of this series form a series of numbers in arithmetical progression; and conversely, the reciprocals of an arithmetical progression are in harmonical proportion.

Recurring Series. A series of which any term is formed by the addition of a certain number of preceding terms, multiplied or divided by any determinate numbers whether positive or negative. Thus 2. 3. 19. 101. 543. 2917. 15671, &c. is a recurring series, each term of which is formed by the addition of the two preceding terms, the first of which being previously multiplied by the constant quantity 2 and the other by 5. Thus the third term $19\; =\; 2\; \backslash times\; 2\; +\; 3\; \backslash times\; 5$; the fourth term $101\; =\; 3\; \backslash times\; 2\; +\; 19\; \backslash times\; 5$, &c.

The principal operation in a series of this nature is that of finding its sum.—For this purpose, the two first and two last terms of the series must be given, together with the constant multipliers.

Let be any number of terms of a series formed according to the above law, each successive term being equal to the sum of the products of the two preceding terms, the first being multiplied by the given quantity $m$, and the other by the given quantity $n$. Hence we will have the following series of equations Then adding these equations, we obtain $c\; +\; d\; +\; e\; +\; f\; =\; m\; \backslash times\; a\; +\; b\; +\; c\; +\; d\; +\; n\; \backslash times\; b\; +\; c\; +\; d\; +\; e$. Now the first member of this equation is the sum of all the terms except the two first; the quantity by which $m$ is multiplied in the second member is the sum of all the terms except the two last; and that by which $n$ is multiplied is the sum of all the terms except the first and last. Now let $s\; =$ sum of the series; then $s\; -\; a\; -\; b\; =\; m\; \backslash times\; s\; -\; c\; -\; f\; +\; n\; \backslash times\; s\; -\; a\; -\; f$

Let the sum of the first seven terms of the above series be required?

Reversion of Series is the method of finding the value of the quantity whose several powers are involved in a series, in terms of the quantity which is equal to the given series.

In order to this, a series must be assumed, which being involved and substituted for the quantity equal to the series, and its powers, neglecting those terms whose powers exceed the highest power to which it is proposed to extend the series.

Let it be required to revert the series ; or, to find $x$ in an infinite series expressed in the powers of $y$.

Substitute $y^1$ for $x$, and the indices of the powers of $y$ in the equation will be 1, 2, 3, 4, &c. and 1, therefore $n=1$; and the differences are 0. 1. 2. 3. 4. 5. &c. Hence, in this case, the series to be assumed is which being involved and substituted for the respective powers of $x$, then we have

Whence, by comparing the homologous terms, we have $aAy\; =\; y$; therefore $A\; =\; \backslash frac\{1\}\{a\},\; B\; =\; \backslash frac\{b\}\{a^2\},\; C\; =\; \backslash frac\{2b^2\; -\; ac\}\{a^3\},\; D\; =\; \backslash frac\{2b^3\; +\; 3c^2\; -\; 2abc\; +\; a^2d\}\{a^4\}$; &c. and consequently

1^st, Let . There $a$ being in this case equal to 1, we shall, by substituting these values, have

2^d, Let ; to find $x$? In this example we have ; whence

3^d, Let to find $x$?

Put $r\; -\; a\; =\; v$; then By comparison we find

P p

Hence

Summation of Series is the method of finding the sum of the terms of an infinite series produced to infinity, or the sum of any number of terms of such a series.

The value of any arithmetical series, as $1^2\; +\; 2^2\; +\; 3^2\; +\; 4^2\; +\; \backslash dots\; +\; n^2$, varies according as ($n$) the number of its terms varies; and therefore, if it can be expressed in a general manner, it must be explicable by $n$ and its powers with determinate coefficients; and those powers, in this case, must be rational, or such whole indices are whole positive numbers; because the progression, being a whole number, cannot admit of surd quantities. Lastly, it will appear that the greatest of the said indices cannot exceed the common index of the series by more than unity: for, otherwise, when $n$ is taken indefinitely great, the highest power of $n$ would be indefinitely greater than the sum of all the rest of the terms.

Thus the highest power of $n$, in an expression exhibiting the value of $1^2\; +\; 2^2\; +\; 3^2\; +\; 4^2\; +\; \backslash dots\; +\; n^2$, cannot be greater than $n^2$; for $1^2\; +\; 2^2\; +\; 3^2\; +\; 4^2\; +\; \backslash dots\; +\; n^2$ is manifestly less than $n^2$, or $n^2\; +\; n^2\; +\; n^2\; +\; \backslash dots$, &c. continued to $n$ terms; but $n^2$, when $n$ is indefinitely great, is indefinitely greater than $n^2$, or any other interior power of $n$, and therefore cannot enter into the equation. This being premised, the method of investigation may be as follows:

1. Required the sum of $n$ terms of the series $1\; +\; 2\; +\; 3\; +\; 4\; +\; \backslash dots\; +\; n$?

Let $A\; n^2\; +\; B\; n$ be assumed, according to the foregoing observations, as an universal expression for the value of $1\; +\; 2\; +\; 3\; +\; 4\; +\; \backslash dots\; +\; n$, where $A$ and $B$ represent unknown but determinate quantities. Therefore, since the equation is supposed to hold universally, whatsoever is the number of terms, it is evident, that if the number of terms be increased by unity, or, which is the same thing, if $n\; +\; 1$ be wrote therein instead of $n$, the equation will still subsist; and we shall have $A\; \backslash times\; n^2\; +\; B\; n\; +\; 1\; +\; B\; \backslash times\; n\; +\; 1\; =\; 1\; +\; 2\; +\; 3\; +\; 4\; +\; \backslash dots\; +\; n\; +\; n\; +\; 1$. From which the first equation being subtracted, there remains $A\; \backslash times\; n^2\; +\; B\; n\; +\; 1\; +\; B\; n\; +\; 1\; =\; n\; +\; 1$; this contracted will be $2\; A\; n\; +\; A\; +\; B\; =\; n\; +\; 1$; whence we have $2\; A\; -\; 1\; \backslash times\; n\; +\; A\; +\; B\; -\; 1\; =\; 0$. Wherefore, by taking $2\; A\; -\; 1\; =\; 0$, and $A\; +\; B\; -\; 1\; =\; 0$, we have $A\; =\; \backslash frac\{1\}\{2\}$, and $B\; =\; \backslash frac\{1\}\{2\}$; and consequently $1\; +\; 2\; +\; 3\; +\; 4\; +\; \backslash dots\; +\; n\; (=\; A\; n^2\; +\; B\; n)\; =\; \backslash frac\{n^2\}\{2\}\; +\; \backslash frac\{n\}\{2\}\; =\; \backslash frac\{n\; \backslash times\; n\; +\; 1\}\{2\}$.

What is the sum of the ten first terms of the series ?

2. Required the sum of the series $1^2\; +\; 2^2\; +\; 3^2\; +\; \backslash dots\; +\; n^2$, or $1\; +\; 4\; +\; 9\; +\; 16\; +\; \backslash dots\; +\; n^2$.

Let $A\; n^2\; +\; B\; n^3\; +\; C\; n$, according to the aforesaid observations, be assumed $=\; 1^2\; +\; 2^2\; +\; 3^2\; +\; \backslash dots\; +\; n^2$; then, as in the preceding case, we shall have $A\; \backslash times\; n^2\; +\; B\; \backslash times\; n^3\; +\; C\; \backslash times\; n\; +\; 1\; =\; 1^2\; +\; 2^2\; +\; 3^2\; +\; \backslash dots\; +\; n^2\; +\; (n\; +\; 1)^2$; that is, by involving $n\; +\; 1$ to its several powers, $A\; n^2\; +\; 3\; A\; n^3\; +\; 3\; A\; n^2\; +\; A\; n\; +\; B\; n^3\; +\; 3\; B\; n^2\; +\; 3\; B\; n\; +\; B\; +\; C\; n\; +\; C\; =\; 1^2\; +\; 2^2\; +\; 3^2\; +\; \backslash dots\; +\; n^2\; +\; (n\; +\; 1)^2$; from which subtracting the former equation, we obtain $3\; A\; n^2\; +\; 3\; A\; n\; +\; A\; +\; 2\; B\; n\; +\; B\; +\; C\; (=\; n\; +\; 1)^2)\; =\; n^2\; +\; 2\; n\; +\; 1$; and consequently $3\; A\; -\; 1\; \backslash times\; n^2\; +\; 3\; A\; +\; 2\; B\; -\; 2\; \backslash times\; n\; +\; A\; +\; B\; +\; C\; -\; 1\; =\; 0$; whence $3\; A\; -\; 1\; =\; 0$, $3\; A\; +\; 2\; B\; -\; 2\; =\; 0$, and $A\; +\; B\; +\; C\; -\; 1\; =\; 0$; therefore $A\; =\; \backslash frac\{1\}\{3\}$, $B\; =\; \backslash frac\{2\; -\; 3\; A\}\{3\}\; =\; \backslash frac\{1\}\{3\}$, $C\; =\; 1\; -\; A\; -\; B\; =\; \backslash frac\{1\}\{3\}$, and consequently $1\; +\; 4\; +\; 9\; +\; 16\; +\; \backslash dots\; +\; n^2\; =\; \backslash frac\{n^2\}\{3\}\; +\; \backslash frac\{n^2\}\{3\}\; +\; \backslash frac\{n^2\}\{3\}$, or $\backslash frac\{n^2\; \backslash times\; n\; +\; 1\; \backslash times\; 2\; n\; +\; 1\}\{6\}$.

What is the sum of the ten first terms of the series ?

3. Required the sum of the series $1^2\; +\; 2^2\; +\; 3^2\; +\; 4^2\; +\; \backslash dots\; +\; n^2$, or $1\; +\; 8\; +\; 27\; +\; 64\; +\; \backslash dots\; +\; n^3$?

By putting $A\; n^2\; +\; B\; n^3\; +\; C\; n^4\; +\; D\; n\; =\; 1\; +\; 8\; +\; 27\; +\; 64\; +\; \backslash dots\; +\; n^3$; and proceeding as above, we shall have $4\; A\; n^2\; +\; 6\; A\; n^3\; +\; 4\; A\; n^4\; +\; A\; n^5\; +\; 3\; B\; n^3\; +\; 3\; B\; n^4\; +\; B\; n^5\; +\; 2\; C\; n\; +\; C\; +\; D\; (=\; n\; +\; 1)^3)\; =\; n^3\; +\; 3\; n^4\; +\; 3\; n^5\; +\; 1$, and therefore $4\; A\; -\; 1\; \backslash times\; n^2\; +\; 6\; A\; +\; 3\; B\; -\; 3\; \backslash times\; n^2\; +\; 4\; A\; +\; 3\; B\; +\; 2\; C\; -\; 3\; \backslash times\; n\; +\; A\; +\; B\; +\; C\; +\; D\; -\; 1\; =\; 0$. Hence $A\; =\; \backslash frac\{1\}\{4\}$, $B\; =\; \backslash left(\; \backslash frac\{3\; -\; 6\; A\}\{3\}\; \backslash right)\; =\; \backslash frac\{1\}\{2\}$, $C\; =\; \backslash left(\; \backslash frac\{3\; -\; 4\; A\; -\; 3\; B\}\{2\}\; \backslash right)\; =\; \backslash frac\{1\}\{4\}$, $D\; =\; 1\; -\; A\; -\; B\; -\; C\; =\; 0$; and therefore $1^2\; +\; 2^2\; +\; 3^2\; +\; 4^2\; +\; \backslash dots\; +\; n^2\; =\; \backslash frac\{n^2\}\{4\}\; +\; \backslash frac\{n^2\}\{2\}\; +\; \backslash frac\{n^2\}\{4\}$, or $\backslash frac\{n^2\; \backslash times\; n\; +\; 1^2\}\{4\}$.

In the very same manner it will be found, that

What is the sum of the ten first terms of the series ?

4. Required the sum of $n$ terms of the series of triangular numbers $0,\; 1,\; 3,\; 6,\; 10,\; \backslash dots\; +\; n$?

Let $A\; n^2\; +\; B\; n^3\; +\; C\; n\; =\; 0,\; 1,\; 3,\; 6,\; 10,\; \backslash dots\; +\; n$. Now the $n\; +\; 1$th term of this series, by Example 2. is $\backslash frac\{n^2\}\{2\}\; +\; \backslash frac\{n\}\{2\}$. Then $A\; \backslash times\; n^2\; +\; B\; \backslash times\; n^3\; +\; C\; \backslash times\; n\; =\; \backslash frac\{n^2\}\{2\}\; +\; \backslash frac\{n\}\{2\}$.

Now, the first equation being subtracted from this, we have $3\; A\; n^2\; +\; 3\; A\; +\; 2\; B\; \backslash times\; n\; +\; A\; +\; B\; +\; C\; =\; \backslash frac\{n^2\}\{2\}\; +\; \backslash frac\{n\}\{2\}$. Or, $3\; A\; n^2\; +\; 3\; A\; n\; +$

Whence, by equating the homologous terms, we have $3A\; =\; \backslash frac\{1\}\{2\}$, and $A\; =\; \backslash frac\{1\}\{6\}$; $\backslash frac\{1\}\{2\}\; -\; 2B\; =\; 3A$; whence $2B\; =\; \backslash frac\{1\}\{2\}\; -\; \backslash frac\{1\}\{2\}\; =\; 0$, $A\; +\; C\; =\; -B$. Hence $C\; =\; -\backslash frac\{1\}\{6\}$. Now, these values being substituted in the above equation, gives the sum $=\; \backslash frac\{n^3\}\{6\}\; -\; \backslash frac\{n\}\{6\}\; =$

By proceeding in the same manner, the sum of $n$ terms of pyramidal numbers, 1, 4, 10, 20, 35, &c.... $n$ will be found $=\; \backslash frac\{n\; \backslash cdot\; n\; +\; 1\; \backslash cdot\; n\; +\; 2\; \backslash cdot\; n\; +\; 3\}\{1\; \backslash cdot\; 2\; \backslash cdot\; 3\; \backslash cdot\; 4\}$. And the sum of any series of figurate numbers is determined by a like formula, the law of continuation being obvious.

What is the sum of the ten first terms of triangular numbers 1, 3, 6, 10, 15, &c.?

5. Let the sum of the series $\backslash frac\{1\}\{R\}\; +\; \backslash frac\{2\}\{R^2\}\; +\; \backslash frac\{3\}\{R^3\}$ continued to $n$ terms be required?

If we multiply this series indefinitely continued by $R\; -\; 1$, or $R^2\; -\; 2R\; +\; 1$, the product is $R$; therefore the amount of the indefinite series is $\backslash frac\{R\}\{R-1\}$, and the sum of $n$ terms may be found by subtracting the terms after the $n$th from that amount. Now, the terms after the $n$th are $\backslash frac\{n+1\}\{R^\{n+1\}\}\; +\; \backslash frac\{n+2\}\{R^\{n+2\}\}$, &c. which may be divided into the two following series:

Now, if we write $a$ for $\backslash frac\{1\}\{R^n\}$, and $r$ for $R\; -\; 1$, and subtract the sum of these two series from the amount of the proposed series indefinitely continued, the remainder will be found $=\; \backslash frac\{1-a\}\{r\}\; \backslash times\; R\; -\; \backslash frac\{na\}\{r\}$.

6. Let the sum of the series $\backslash frac\{n-1\}\{nR\}\; +\; \backslash frac\{n-2\}\{nR^2\}\; +\; \backslash frac\{n-3\}\{nR^3\}$ &c. be required?

This series is equal to the difference of the two following.

The difference of these series is $\backslash frac\{1-a\}\{r\}\; -\; \backslash frac\{1-a\}\{n\}\; \backslash times\; \backslash frac\{1-a\}\{r\}\; +\; \backslash frac\{a\}\{r\}$ Seringapatam.
which reduced becomes $\backslash frac\{n+a-1\; \backslash times\; r\; +\; a-1\}\{nr^2\}$.

To proceed further would lead us far beyond the limits assigned for this article; we must therefore refer those who require more information on this subject to the following authors.—Bertrand's Developpement, &c. vol. 1; Dodion's Mathematical Repository, vol. 1; Emerson's Algebra; Appendix to Gravelend's Algebra; Hutton's Paper on Cubic Equations and Infinite Series, in the Philosophical Transactions for 1780; MacLaurin's Fluxions; Malcolm's Arithmetic; Mafer's Annuities; and Scriptores Logarithmici, &c.; De Moivre's Doctrine of Chances, and a Paper by the same author in the Philosophical Transactions, n^o 240; Simpson's Algebra, Essays, Fluxions, and Miscellanies; Sterling's Summatio et Interpolatio Serierum; Syntagma Matheseos, &c.