1-x-a1-xx-a1-a1a2- -xx-a1x-a2-x-an-a1a2-an-

The correct option is A (x+a_1)(x+a_2)...(x+a_n 1)a_1a_2...a_n 1a_n{x^2+ a_nx+a_n} 1+ xa_1+x(x+a_1)a_1a_2 + ...+ x(x+a_1)(x+a_2)...(x+a_n)a_1a_2...a_n ...

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Byju's Answer

Standard XII

Mathematics

Combination

1+x/a1+xx+a1/...

Question

$1 + \frac{x}{a_{1}} + \frac{x (x + a_{1})}{a_{1} a_{2}} + \dots + \frac{x (x + a_{1}) (x + a_{2} .) . \cdot . \dots \dots \cdot (x + a_{n})}{a_{1} a_{2} . . . a_{n}} =$

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

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

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

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none of these

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Solution

The correct option is A $\frac{(x + a_{1}) (x + a_{2}) . . . (x + a_{n - 1})}{a_{1} a_{2} . . . a_{n - 1} a_{n}} {x^{2} + a_{n} x + a_{n}}$
$1 + \frac{x}{a_{1}} + \frac{x (x + a_{1})}{a_{1} a_{2}} + . . . + \frac{x (x + a_{1}) (x + a_{2}) . . . (x + a_{n})}{a_{1} a_{2} . . . a_{n}}$
$= \frac{a_{1} + x}{a_{1}} + \frac{x (x + a_{1})}{a_{1} a_{2}} + . . . + \frac{x (x + a_{1}) (x + a_{2}) . . . (x + a_{n})}{a_{1} a_{2} . . . a_{n}}$
$= \frac{(x + a_{2}) (x + a_{1})}{a_{1} a_{2}} + . . . + \frac{x (x + a_{1}) (x + a_{2}) . . . (x + a_{n})}{a_{1} a_{2} . . . a_{n}}$
$= \frac{(x + a_{1}) (x + a_{2}) . . . (x + a_{n - 1})}{a_{1} a_{2} . . . a_{n - 1}} + \frac{x (x + a_{1}) (x + a_{2}) . . . (x + a_{n})}{a_{1} a_{2} . . . a_{n}}$
$= \frac{(x + a_{1}) (x + a_{2}) . . . (x + a_{n - 1})}{a_{1} a_{2} . . . a_{n - 1}} {1 + \frac{x (x + a_{n})}{a_{n}}}$
$= \frac{(x + a_{1}) (x + a_{2}) . . . (x + a_{n - 1})}{a_{1} a_{2} . . . a_{n - 1} a_{n}} {x^{2} + a_{n} x + a_{n}}$

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Similar questions

Q. 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}

Q. If

a_{1}, a_{2}, a_{3} \dots \dots \dots, a_{n}

are in

H . P .

then

\frac{a_{1}}{a_{2} + a_{3} + \dots \dots \dots \dots + a_{n}}, \frac{a_{2}}{a_{1} + a_{3} + \dots \dots \dots + a_{n}} \dots \dots \frac{a_{n}}{a_{1} + a_{2} + \dots \dots + a_{n - 1}}

are in

Let $a_{1}, a_{2}, . . . . ., a_{n}$ be fixed real numbers such that $f (x) = (x - a_{1}) (x - a_{2}) . . . . (x - a_{n})$ What is $l i m_{x \to a_{1}}$ f(x)? For $a \neq a_{1}, a_{2}, . . . ., a_{n}$ compute $l i m_{x \to a} f (x) .$

A_{i} = \frac{x - a_{i}}{| x - a_{i} |}, i = 1, 2, . . . . ., n,

and

a_{1} < a_{2} < a_{3} < . . . . . < a_{n} .

1 \leq m \leq n, m \in N,

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Q. If

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1+xa1+x(x+a1)a1a2+…+x(x+a1)(x+a2.).⋅.⋯⋯⋅(x+an)a1a2...an=

$1 + \frac{x}{a_{1}} + \frac{x (x + a_{1})}{a_{1} a_{2}} + \dots + \frac{x (x + a_{1}) (x + a_{2} .) . \cdot . \dots \dots \cdot (x + a_{n})}{a_{1} a_{2} . . . a_{n}} =$