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Now, if =2" be resolved into a series, it is manifest

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x-y be resolved into a x-y

series, the first and last terms will be x-1, and y-1 repectively.

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Multiply by (1-z,) and transpose 1-2", whence,

1+az+bz2+cz3, pzn-2+qz"-1,
—z—az3—bz3, —mzn-2-pzn—1—qz",

-1

+

=0.

.. (art. 89,) a-1-0, b-a-0, c-b=0, p-m=0, q-p=0, 1-q=0.

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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

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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.

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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

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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

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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",

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Substituting these expressions in equations A and B, and subtracting, we have

vm—wm—a(x—y)+b(x2—y3)+c(x3—y3)+
d.(x1-y1)+, &c.

.. (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

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=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,)

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Multiply by v"=1+x, and we have

-.(1+x)ñ=a+2bx+3cx2+4dx2+5ex2, &c.

m

m

m

m

n

n

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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

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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.}

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