### Part I. Section III. Chapter 4. “Of the Summation of Arithmetical Progressions.”

412 It is often necessary also to find the sum of an arithmetical progression. This might be done by adding all the terms together; but as the addition would be very tedious, when the progression consisted of a great number of terms, a rule has been devised, by which the sum may be more readily obtained.

413 We shall first consider a particular given progression, such that the first term is 2, the difference 3, the last term 29, and the number of terms 10;

 1 2 3 4 5 6 7 8 9 10 2, 5, 8, 11, 14, 17, 20, 23, 26, 29

In this progression we see that the sum of the first and last term is 31; the sum of the second and the last but one 31; the sum of the third and the last but two 31, and so on: hence we conclude, that the sum of any two terms equally distant, the one from the first, and the other from the last term, is always equal to the sum of the first and the last term.

414 The reason of this may be easily traced; for if we suppose the first to be $$a$$, the last $$z$$, and the difference $$d$$, the sum of the first and the last term is $$a+z$$; the second term being $$a + d$$, and the last but one $$z-d$$, the sum of these two terms is also $$a+z$$. Farther, the third time being $$a + 2d$$, and the last but two $$z-2d$$, it is evident that these two terms also, when added together, make $$a+z$$ and the demonstration may be easily extended to any other two terms equally distant from the first and last.

415 To determine, therefore, the sum of the progression proposed, let us write the same progression term by term, inverted, and add the corresponding terms together, as follows:

 2 +5 +8 +11 +14 +17 +20 +23 +26 +29 29 +26 +23 +20 +17 +14 +11 +8 +5 +2 31 31 31 31 31 31 31 31 31 31

This series of equal terms is evidently equal to twice the sum of the given progression: now, the number of those equal terms is 10, as in the progression, and their sum consequently is equal to 10 · 31 = 310. Hence, as this sum is twice the sum of the arithmetical progression, the sum required must be 155.

416 If we proceed in the same manner with respect to any arithmetical progression, the first term of which is $$a$$, the last $$z$$, and the number of terms $$n$$; writing under the given progression the same progression inverted, and adding term to term, we shall have a series of $$n$$ terms, each of which will be expressed hy $$a + z$$; therefore the sum of this series will be $$n(a+z)$$, which is twice the sum of the proposed arithmetical progression; the latter, therefore, will be represented by

$\frac{n(a+z)}{2}.$

417 This result furnishes an easy method of finding the sum of any arithmetical progression ; and inay be reduced to the following rule:

Multiply the sum of the first and the last term by the number of terms, and half the product will be the sum of the whole progression. Or, which amounts to the same, multiply the sum of the first and the last term by half the number of terms. Or, multiply half the sum of the first and the last term by the whole number of terms.

418 It will be necessary to illustrate this rule by some examples.

First, let it be required to find the sum of the progression of the natural numbers, 1, 2, 3, etc. to 100. This will be, by the first rule, $$\frac{100\cdot 101}{2} =\frac{10100}{2}= 5050$$.

If it were required to tell how many strokes a clock strikes in twelve hours; we must add together the numbers 1, 2, 3, as far as 12; now this sum is found immediately to be $$\frac{12\cdot 13}{2} = 6\cdot 13=78$$. If we wished to know the sum of the same progression continued to 1000, we should find it to be 500500; and the sum of this progression, continued to 10000, would be 50005000.

419 Suppose a person buys a horse, on condition that for the first nail he shall pay 5 pence, for the second 8 pence, for the third 11 pence, and so on, always increasing 3 pence more for each nail, the whole number of which is 32; required the purchase of the horse?

In this question it is required to find the sum of an arithmetical progression, the first term of which is 5, the difference 3, and the number of terms 32; we must therefore begin by determining the last term; which is found by the rule, in Art. 406 and Art. 411, to be 5 + (31·3) = 98; after which the sum required is easily found to be $$\frac{103\cdot 32}{2}=103\cdot 16$$; whence we conclude that the horse costs 1648 pence, or 6l. 17s. 4d.

420 Generally, let the first term be $$a$$, the difference $$d$$, and the number of terms $$n$$ ; and let it be required to find, by means of these data, the sum of the whole progression. As the last term must be $$a + (n - 1)d$$, the sum of the first and the last will be $$2a + (n-1)d$$; and multiplying this sum by the number of terms $$n$$, we have $$2na + n(n - l)d$$; the sum required therefore will be

$na + \frac{n(n-1)d}{2}.$

Now, this formula, if applied to the preceding example, or to $$a = 5$$, $$d = 3$$, and $$n = 32$$, gives

$5\cdot 32+\frac{32\cdot 31\cdot 3}{2} = 160+1488=1648;$

the same sum that we obtained before.

421 If it be required to add together all the natural numbers from 1 to $$n$$, we have, for finding this sum, the first term 1, the last term $$n$$, and the number of terms $$n$$; therefore the sum required is $$\frac{n^2+n}{2}=\frac{n(n+1)}{2}$$. If we make $$n=1766$$, the sum of all the numbers, from 1 to 1766, will be 883, or half the number of terms, multiplied by 1767, =1560261.

422 Let the progression of uneven numbers be proposed, 1, 3, 5, 7, etc. continued to $$n$$ terms, and let the sum of it be required. Here the first term is 1, the difference 2, the number of terms $$n$$; the last term will therefore be $$1+(n-1) 2=2n-1$$, and consequently the sum required $$=n^2$$.

The whole therefore consists in multiplying the number of terms by itself; so that whatever number of terms of this progression we add together, the sum will be always a square, namely, the square of the number of terms; which we shall exemplify as follows

 Indices 1 2 3 4 5 6 7 8 9 10 etc. Progression 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, etc. Sum 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, etc.

423 Let the first term be 1, the difference 3, and the number of terms $$n$$; we shall have the progression 1, 4, 7, 10, etc. the last term of which will be $$1+(n-3)3=3n-2$$; wherefore the sum of the first and the last term is $$3n-1$$, and consequently the sum of this progression is equal to $$\frac{n(3n-1)}{2}=\frac{3n^2-n}{2}$$; and if we suppose $$n = 20$$, the sum will be 10 · 59=590.

424 Again, let the first term be 1, the difference $$d$$, and the number of terms $$n$$; then the last term will be $$1 + (n - 1)d$$; to which adding the first, we have $$2 + (n - 1)d$$, and multiplying by the number of terms, we have $$2n+n(n-1)d$$; whence we deduce the sum of the progression $$n+\frac{n(n-1)d}{2}$$.

And by making $$d$$ successively equal to 1, 2, 3, 4, etc., we obtain the following particular values, as shown in the subjoined Table.

Difference Sum
$$d=1$$ $$n+\dfrac{n(n-1)}{2}$$ $$=\dfrac{n^2+n}{2}$$
$$d=2$$ $$n+\dfrac{2n(n-1)}{2}$$ $$=n^2$$
$$d=3$$ $$n+\dfrac{3n(n-1)}{2}$$ $$=\dfrac{3n^2-n}{2}$$
$$d=4$$ $$n+\dfrac{4n(n-1)}{2}$$ $$=2n^2-n$$
$$d=5$$ $$n+\dfrac{5n(n-1)}{2}$$ $$=\dfrac{5n^2-3n}{2}$$
$$d=6$$ $$n+\dfrac{6n(n-1)}{2}$$ $$=3n^2-2n$$
$$d=7$$ $$n+\dfrac{7n(n-1)}{2}$$ $$=\dfrac{7n^2-5n}{2}$$
$$d=8$$ $$n+\dfrac{8n(n-1)}{2}$$ $$=4n^2-3n$$
$$d=9$$ $$n+\dfrac{9n(n-1)}{2}$$ $$=\dfrac{9n^2-7n}{2}$$
$$d=10$$ $$n+\dfrac{10n(n-1)}{2}$$ $$=5n^2-4n$$

#### Editions

1. Leonhard Euler. Elements of Algebra. Translated by Rev. John Hewlett. Third Edition. Longmans, Hurst, Rees, Orme, and Co. London. 1822.
2. Leonhard Euler. Vollständige Anleitung zur Algebra. Mit den Zusätzen von Joseph Louis Lagrange. Herausgegeben von Heinrich Weber. B. G. Teubner. Leipzig and Berlin. 1911. Leonhardi Euleri Opera omnia. Series prima. Opera mathematica. Volumen primum.