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Derivative from First Principle

If 1+a+a2+a...

Question

If $1 + a + a^{2} + a^{3} + . . . . . . . . . + a^{n} = (1 + a) (1 + a^{2}) (1 + a^{4})$ then $n$ is given by

A

3

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B

5

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C

7

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D

9

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Solution

The correct option is C $7$
Given,
$1 + a + a^{2} + a^{3} + . . . . . . . . . + a^{n} = (1 + a) (1 + a^{2}) (1 + a^{4})$
or, $1 + a + a^{2} + a^{3} + . . . . . . . . . + a^{n} = (1 + a + a^{2} + a^{3}) (1 + a^{4})$
or, $1 + a + a^{2} + a^{3} + . . . . . . . . . + a^{n} = (1 + a + a^{2} + a^{3} + a^{4} + a^{5} + a^{6} + a^{7})$
Comparing both sides we get, $n = 7$ .

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Similar questions

Q. Let the sequence

a_{1}, a_{2},

a_{n}

form an A.P. and let

a_{1} = 0

\frac{a_{3}}{a_{2}} + \frac{a_{4}}{a_{3}} + \frac{a_{5}}{a_{4}} +

+ \frac{a_{n}}{a_{n - 1}} - a_{2} (\frac{1}{a_{2}} + \frac{1}{a_{3}} + . . . + \frac{1}{a_{n - 2}}) = \frac{a_{n - 1}}{a_{2}} + \frac{a_{2}}{a_{n - 1}}

A_{1} A_{2} A_{3} . . . A_{n}

be a regular polygon of

n

\frac{1}{A_{1} A_{2}} = \frac{1}{A_{1} A_{3}} + \frac{1}{A_{1} A_{4}},

a_{r} > 0; \forall r, n \in N

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

are in A.P, then

\frac{a_{1} + a_{2 n}}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{a_{2} + a_{2 n - 1}}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \frac{a_{3} + a_{2 n - 2}}{\sqrt{a_{3}} + \sqrt{a_{4}}} + . . . + \frac{a_{n} + a_{n + 1}}{\sqrt{a_{n}} + \sqrt{a_{n + 1}}} =

a_{1}, a_{2}, a_{3}, a_{4}, a_{5} > 0

\frac{a_{1}}{a_{2} + a_{3}} + \frac{a_{2}}{a_{3} + a_{4}} + \frac{a_{3}}{a_{4} + a_{5}} + \frac{a_{4}}{a_{5} + a_{1}} + \frac{a_{5}}{a_{1} + a_{2}} \geq \frac{5}{2}

a_{1}, a_{2}, a_{3}, a_{4}

are in AP, then

\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \frac{1}{\sqrt{a_{3}} + \sqrt{a_{4}}}

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