Question

Hay brainly users solve this problem If the ratio between a and b is
$\frac{a {}^{n + 1} + b {}^{n + 1} }{a {}^{n} + b {}^{n} }$
mean then a value of n is known ?

Answer 1

Hay dear user -

we know that formula for GP => √ab

so,

$\sqrt{ab} = \frac{a {}^{n + 1} + b {}^{n + 1} }{a {}^{n} + b {}^{n} }$

if we square both sides then we can get

${ab} = \frac{(a {}^{n + 1} + b {}^{n + 1} ) {}^{2} }{(a {}^{n} + b {}^{n} ) {}^{2} }$


$ab(a {}^{2n} + b {}^{2n} + 2a {}^{n} b {}^{n} ) = a {}^{2n +2 + } + b {}^{2n + 2} + 2a {}^{n + 1} b {}^{n + 1}$


$aba {}^{2n} + abb {}^{2n} + 2a {}^{n} b {}^{n} \times ab = a {}^{2n + 1} \times a + bb {}^{2n + 1} + 2a {}^{n + 1} b {}^{n + 1}$



$ba {}^{2n + 1} + ab {}^{2n + 1} + 2a {}^{n + 1} b {}^{n + 1} = aa {}^{2n + 1} + bb {}^{2n + 1} + 2a {}^{n + 1} b {}^{n + 1}$



$ba {}^{2n + 1} - aa {}^{2n + 1} = bb {}^{2n + 1} - ab {}^{2n + 1}$



$a {}^{2n + 1} (b - a) = b {}^{2n + 1}(b - a)$



$\frac{a {}^{2n + 1} }{b {}^{2n + 1} } = 1$

if we substitute both powers then,

$( \frac{a}{b} ) {}^{2n + 1} = ( \frac{a}{b} ) {}^{0}$



$2n + 1 = 0 \\ \\ 2n = - 1 \\ \\ n = - \frac{1}{2}$

Answer 2

This is not mathematics