Show that 1 a 0 - x a 0 a 1 - x 2 a 0 a 1 a 2 - - x n a 0 a 1 a 2 - a n - 1 a 0 - a 0 x a 1 -x- a 1 x a 2 -x- - a n-1 x a n -x-

⇒ Put 1 a _ r + x a _ r a _ r+1 = 1 a _ r +y_ r then ( a _ r + y _ r ) ( a _ r+1 +x ) = a _ r a _ r+1 ; y _ r = a _ r x a _ r+1 ...

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Sum of n Terms

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Question

Show that
$\frac{1}{a_{0}} + \frac{x}{a_{0} a_{1}} + \frac{x^{2}}{a_{0} a_{1} a_{2}} + \dots + \frac{x^{n}}{a_{0} a_{1} a_{2} \dots a_{n}} = \frac{1}{a_{0} -} \frac{a_{0} x}{a_{1} + x -} \frac{a_{1} x}{a_{2} + x -} \dots \frac{a_{n - 1} x}{a_{n} + x}$ .

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\frac{1}{a_{0}} - \frac{x}{a_{0} a_{1}} + \frac{x^{2}}{a_{0} a_{1} a_{2}} - \dots + {(- 1)}^{n} \frac{x^{n}}{a_{0} a_{1} a_{2} \dots a_{n}}

{(1 + x + x^{2})}^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{2 n} x^{2 n},

a_{0} + a_{1} + a_{2} + \dots . + a_{2 n}

a_{0} - a_{1} + a_{2} - a_{3} \dots . . + a_{2 n}

n

{(1 + x)}^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{n} x^{n}

a_{0} + a_{1} + a_{2} + \dots + a_{n}

(1 + x + x^{2} + x^{3} + \dots \dots . x^{p})^{n} = a_{0} + a_{1^{X}} + a_{2} x^{2} + \dots . . + a_{n p} x^{n p}

a_{0} + a_{1} + a_{2} + \dots . + a_{n p} =

x + 1

f (x) = a_{0} x^{n} + a_{1} x^{n - 1} + a_{2} x^{n - 2} + \dots + a_{n} = 0

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