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Higher Order Derivatives

Let 1+x221+xn...

Question

Let $(1 + x^{2})^{2} (1 + x)^{n} = A_{0} + A_{1} x + A_{2} x^{2} + . . . . . . . . . . . . . . I f A_{0}, A_{1}, A_{2}$ are in A.P., then the value of n is

A

2

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B

3

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C

5

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D

7

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Solution

The correct options are
A
2

B
3

$(1 + x^{2})^{2} (1 + x)^{n} = (1 + 2 x^{2} + x^{4})^{n} C_{0} +^{n} C_{1} x +^{n} C_{2} x^{2} + . . . . . . . . . . . . . .)$

$=^{n} C_{0} +^{n} C_{1} x + (^{n} C_{2} + {2.}^{n} C_{0}) x^{2} + . . . . . .$

Hence $A_{0} = 1; A_{1} =^{n} C_{1}; A_{2} =^{n} C_{2} + 2$ which are in A.P.

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Similar questions

{(1 + x^{2})}^{2} {(1 + x)}^{n} = A_{0} + A_{1} x + A_{2} X^{2} + \cdot \cdot \cdot

A_{0}, A_{1}, A_{2}

are in A.P. then the value of n is

{(1 + x^{2})}^{2} {(1 + x)}^{n} = A_{0} + A_{1} x + A_{2} x^{2} + . . . .

A_{0}, A_{1}, A_{2}

are in A.P, then the value of

n

n \in N

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

a_{0} + a_{1} + a_{2} . . . . a_{n - 1}

{(1 + x)}^{n} = 1 + a_{1} x + a_{1} x^{2} + . . . . + a_{n} x^{n}

a_{1}

a_{2}

a_{3}

are in A.P., then the value of n is

(1 - x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . + a_{2 n} x^{2 n}

a_{0}, a_{1}, a_{2}, . . ., a_{2 n}

are in A.P. then prove that

a_{n} = \frac{1}{2 n + 1}

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