XLIII. The principle of summing these senes may be proved generally as follows: Let 1, a, b, c, d......I be a series of any order, such that the sum of ʼn terms may be found by multiplying the (+1, th term by n, and dividing the product by m. If l is the (n+1) th term, and s the sum of all the terms, we shall have by hypothesis That is, n I will be m times the sum of the series. The next higher series will be formed from this as follows: 1+a+b+c+d+....k + l = (n + 1)th. first term 1 of the original series 1, a, b, &c., forms the of the new series; the sum of the first two forms the the sum of the first three forms the third term, ms forms the (n+1)th term. term, be written n And let a line be the first row, the e nth row may be By the rule given above for finding the sum of the series 1, 3, 6, 10, &c. The sum of four terms, or 1+3+6+10= The sum of five terms, or 1+3+6+10+15= The five 21s are 3 times The four 15s are 3 times and so of the rest. 1+3+6+10 + 15. 1+3+6+10 It is easy to see that this principle will extend to any number of terms. Therefore to find the sum of n terms of the series 1, 4, 10, 20, &c., multiply the (n+1)th term of the series by n, and divide the product by 4, and the quotient will be the sum required. But the (n+1)th term of this series is equal to the sum of (n + 1) terms of the preceding series. The nth term of the preceding series being This being multiplied by n and divided by 4, gives ¿'ll − n (n + 1) (n + 2) (n+3) 1 X 2 X 3 X 4 XLIII. The principle of summing these series may be proved generally as follows: Let 1, a, b, c, d . . . . . .l be a series of any order, such that the sum of ʼn terms may be found by multiplying the (n + 1)th term by n, and dividing the product by m. If l is the (n + 1) th term, and s the sum of all the terms, we shall have by hypothesis That is, n I will be m times the sum of the series. The next higher series will be formed from this as follows: =nth 1+a+b+c+d 1+a+b+c+d+....k 1+a+b+c+ d + .... k + l = (n + 1)th. The first term 1 of the original series 1, a, b, &c., forms the first term of the new series; the sum of the first two forms the second term; the sum of the first three forms the third term, &c., and the sum of (n + 1) terms forms the (n + 1)th term. Let the series forming the (n + 1)th term, be written n times, one under the other, term for term. And let a line be drawn diagonally, so that the first term of the first row, the first two of the second row, and n terms of the nth row may be at the left, and below the line. The The coefficient of the first term of every power is 1. coefficient of the second term of every power is formed by adding together the coefficients of the first and second terms of the preceding power. The coefficient of the third term of every power is formed by adding together the coefficients of the second and third terms of the preceding power. The coefficient of the fourth term of every power is found by adding together the coefficients of the third and fourth terms of the preceding power. And so of the rest. This law, though perhaps sufficiently evident by inspection, may be easily demonstrated. Suppose the above law to hold true as far as some power which we may designate by n. The literal part of the nth power will be formed thus. The We cannot write all the terms without assigning a particular value to n. We can write a few of the first and last. points between show that the number of terms is indeterminate; there may or may not be more than are written. Suppose that A is the coefficient of the second term, B that of the third, &c. and let the whole be multiplied by a +x, which will produce the next higher power, or the (n+1)th power. |