Page:Elementary algebra (1896).djvu/364

346 ALGEBRA. Let the rth and (r+1)th be any two consecutive terms. The (r+1)th term is obtained by multiplying the rth term by ; that is,  Art. 410.

The factor decreases as r increases; hence the (r+1)th term is not always greater than the rth term, but only until becomes equal to 1, or less than 1.

Now, so long as ;

that is,

If (n+1)b a+b be an integer, denote it by p; then if r=p the multiplying factor becomes 1, and the (p+ 1)th term is equal to the pth; and these are greater than any other term.

If be not an integer, denote its integral part by q; then the greatest value of r consistent with (1) is q; hence the (q+1)th term is the greatest.

Since we are only concerned with the numerically greatest term, the investigation will be the same for (a-b)n; therefore in any numerical example it is unnecessary to consider the sign of the second term of the binomial. Also it will be found best to work each example independently of the general formula.

Ex. Find the greatest term in the expansion of (1 + 4x)8, when x has the value 1 3.

Denote the ith and (r+1)th terms by Tr; and Tr+1: respectively ; then

hence Tr+1> Tr, so long as --- x 37 1;

that is, 36-4r>3r, or 36>7r.