Now, if =2" be resolved into a series, it is manifest x-y be resolved into a x-y series, the first and last terms will be x-1, and y-1 repectively. Multiply by (1-z,) and transpose 1-2", whence, 1+az+bz2+cz3, pzn-2+qz"-1, -1 + =0. .. (art. 89,) a-1-0, b-a-0, c-b=0, p-m=0, q-p=0, 1-q=0. xn−1+xn−2y+xn¬3y2+xn−4y3, xn−n—1yn−2+yn−1 ̧ Whence, xy"—(x—y)×{x~~1+xn−2y+x2¬3y3. yn-1.} Here the indices of x descending regularly from n-1, to 0, or n-n, it appears the number of terms in the series is n. 91. Required to develope (1+x)" in a series, n being a whole positive number. It is easy to perceive that the first term of the series is 1. Assume then, (1+x)"=1+ax+bx2+cx2+dx2+, &c. (A.) Consequently, (1+y)"=1+ay+by3+cy3+dya, &c. By subtraction, (1+x)”—(1+y)”—a(x—y)+b(x2—y3) +c(x3—y3)+d(xa—y1,) &c. Divide by 1+2-1+y=x-y, and we shall have Or (1+x)n-1+(1+x)n−2×1+y(1+y)"−1 =a+b(x+y)+c(x2+xy+y3) +d(x3+x3y+xy3+y3)+, &c. This equation being true, whatever value may be assigned to x and y, let us suppose x=y, then this equa tion becomes (art. 90,) n.(1+x)-1=a+2bx+3cx2+4dx+5ex4, &c. Multiply by (1+x) ·. n.(1+x)"=a+2bx+3cx2+4dx3+5ex1, &c. But (eq. A.) ax+2bx2+3cx3+4dx+, &c. n.(1+x)"=n+nax+nbx2+ncx3+ndxa, &c. When n is a whole positive number, as here supposed, the series is finite; the number of terms being n+1; for the co-efficient of term n+2 is ers for +6 and its powers. If.(a-b)" be required, we may put-b and its pow n-1 Whence, (a-b)"=a"—na"-1b+n. -an 2f2 2 92. Required to develope (1+x) in a series. m Assume (1+x)=1+ax+bx2+cx3+dx2+ex3,&c.(A.) m Then (1+y)=1+ay+by3+cy3+dy*+ey3, &c. (B.) L Put (1+x)"=v, (1+y)=w, then 1+x=v",1+y=w", Substituting these expressions in equations A and B, and subtracting, we have vm—wm—a(x—y)+b(x2—y3)+c(x3—y3)+ .. (art. 90,) (vw)×{vm-1+vm-2w. wm-1}= (x—y). {a+b(x+y)+c.(x2+xy+y3)+ d(x2+x3y+xy2+y3), &c.} But x-y=1+x-1+y=v"—w”—(v—w) × {v-1+vn-w+. wn-1} .. Dividing the last equation by this, we have vm−1+vm−2w+vm¬3w3. wm-1 =a+b(x+y)+ c(x2+xy+y3)+d(x3+x3y+xy2+y3), &c. Now, suppose x=y, whence v=w, and our equation becomes (art. 90,) Multiply by v"=1+x, and we have -.(1+x)ñ=a+2bx+3cx2+4dx2+5ex2, &c. m m m m n n an.{1+ = m b m α m mnm-2n b3 &c.} m b m m -n b2 n α n 2n a2 n 2n Sn a 93. In the preceding investigation m has been supposed to be affirmative: if we now take m negative, and make the same assumptions as before, we have 1 vm.wm (v—w).{vm−1+vm-2w+vm-sw3. wm-1}= (x—y).{a+b(x+y)+c.(x2+xy+y3)+ d(x3+x3y+xy3+y3)+, &c.}= (v-w).{vn-1+vπ-2w. wn-1} {a+b(x+y)+c(x2+xy+y3)+, &c.} |