contains x', it can be made greater than any finite quantity by taking r sufficiently large; in which case there is no limit to the value of the above series; and therefore the expansion of (1 + x)" as an infinite series in ascending powers of x has no meaning arithmetically intelligible when x is greater than 1. 184. We may remark that we can always expand (x + y)" by the Binomial Theorem; for we may write the expression in either of the two following forms : and we obtain the expansion from the first or second of these according as x is greater or less than y. 185. To find in its simplest form the general term in the expansion of (1− x)−". From this it appears that every term in the expansion of (1-x)" is positive. Although the general term in the expansion of any binomial may always be found as explained in Art. 182, it will be found more expeditious in practice to use the above form of the general term in all cases where the index is negative, retaining the form Example. Find the general term in the expansion of 1 3/1-3x If the given expression had been (1+3x) we should have used the same formula for the general term, replacing 3x by - 3x. 186. The following expansions should be remembered: (1 − x) ̄1 = 1 + x + x2 + x2 + + x2 + ...... ...... 187. The general investigation of the greatest term in the expansion of (1+x)", when n is unrestricted in value, will be found in Art. 189; but the student will have no difficulty in applying to any numerical example the method explained in Art. 172. Example. Find the greatest term in the expansion of (1+x)-" when x= and n= 20. 2 3' Hence for all values of r up to 37, we have Tr+1>Tr; but if r=38, then Tr+1=T, and these are the greatest terms. Thus the 38th and 39th terms are equal numerically and greater than any other term. 188. Some useful applications of the Binomial Theorem are explained in the following examples. Example 1. Find the first three terms in the expansion of Expanding the two binomials as far as the term containing x2, we have If in this Example x= ='002, so that x2=000004, we see that the third term is a decimal fraction beginning with 5 ciphers. If therefore we were required to find the numerical value of the given expression correct to 5 places of decimals it would be sufficient to substitute ⚫002 for x in 1+1 ing the term involving x2. 13 6 x, neglect Example 2. When x is so small that its square and higher powers may be neglected, find the value of Since x2 and the higher powers may be neglected, it will be sufficient to retain the first two terms in the expansion of each binomial. Therefore |