What is the geometric mean of the sequence $1, 2, 4, 8, . . . . . . 2^{n}$ ?

2^{n / 2}

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2^{(n + 1) / 2}

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2^{(n + 1)} - 1

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2^{(n - 1)}

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Solution

The correct option is C $2^{n / 2}$
Given:
$G . M . = {({1.2.4.8......2}^{n})}^{\frac{1}{n + 1}}$

$G . M . = {(2^{0} {.2}^{1} {.2}^{2} {.2}^{3} . . . . . {.2}^{n})}^{\frac{1}{n + 1}}$

$G . M . = (2^{0 + 1 + 2 + 3 + . . . + n})^{\frac{1}{n + 1}}$

$G . M . = {((2)^{n (n + 1) / 2})}^{\frac{1}{n + 1}}$

$∴ G . M . = 2^{n / 2}$

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Similar questions

Simplify 16*2n+1-4*2n/16*2n/1*2n+2-2*2n+1

1, 2, 4, 8, \dots 2^{n}

1 - 2 n + \frac{2 n (2 n - 1)}{2!} - \frac{2 n (2 n - 1) (2 n - 1)}{3!} + . . . + (- 1)^{n - 1} \frac{2 n (n - 1) . . . (n + 2)}{(n - 1)}

= (- 1)^{n + 1} \frac{(2 n)}{2 (n!)^{2}},

^{2 n} C_{1} +^{2 n} C_{3} +^{2 n} C_{5} + . . . . . +^{2 n} C_{n - 1} = 2^{2 n - 1}

^{2 n} C_{1} +^{2 n} C_{3} +^{2 n} C_{5} + . . . . . +^{2 n} C_{n - 1} = 2^{2 n - 2}

a, a_{1}, a_{2}, . . . . ., a_{2 n}, b

a, g_{1}, g_{2}, . . . . g_{2 n}, b

h

a

b

\frac{a_{1} + a_{2 n}}{g_{1} g_{2 n}} + \frac{a_{2} + a_{2 n - 1}}{g_{2} g_{2 n - 1}} + . . . . . + \frac{a_{n} + a_{n + 1}}{g_{n} g_{n + 1}}

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