Question

Let r be a positive number satisfying r^1/1234) + (r^-1/1234) = 2 ..then find (r^4321)+(r^-4321)??​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

We are aware of the identity that :

$\sf{ \: {(a - b)}^{2} = {a}^{2} - 2ab + {b}^{2} \: }$

GIVEN

$\displaystyle \sf{ {r}^{ \frac{1}{1234} }+ {r}^{ \frac{ - 1}{1234} }\: \: = 2}$

TO DETERMINE

$\displaystyle \sf{ {r}^{ 4321 }+ {r}^{ - 4321 }\: \: }$

CALCULATION

Let

$\displaystyle \sf{ x = {r}^{ \frac{1}{1234} } \: \: }$

Then

$\displaystyle \sf{ \frac{1}{x} = {r}^{ \frac{ - 1}{1234} }\: }$

Now

$\displaystyle \sf{ {r}^{ \frac{1}{1234} }+ {r}^{ \frac{ - 1}{1234} }\: \: = 2}$

$\displaystyle \implies \sf{x + \frac{1}{x} = 2}$

$\displaystyle \implies \sf{ \frac{ {x}^{2} + 1 }{x} = 2}$

$\displaystyle \implies \sf{ {x}^{2} + 1 = 2x}$

$\displaystyle \implies \sf{ {x}^{2} - 2x + 1 =0}$

$\displaystyle \implies \sf{ {(x - 1)}^{2} =0}$

$\displaystyle \implies \sf{ x - 1 = 0}$

$\displaystyle \implies \sf{ x = 1}$

$\displaystyle \implies \sf{ \displaystyle \sf{ {r}^{ \frac{1}{1234} } = 1\: \: }}$

$\implies \: \displaystyle \sf{ r = 1 \: \: }$

Hence

$\displaystyle \sf{ {r}^{ 4321 }+ {r}^{ - 4321 }\: \: }$

$= \displaystyle \sf{ {(1)}^{ 4321 }+ {(1)}^{ - 4321 }\: \: }$

$\displaystyle \sf{ = 1 + 1\: \: }$

$= 2$

RESULT

$\boxed{ \displaystyle \sf{ \: \: \: {r}^{ 4321 }+ {r}^{ - 4321 }\: = 2 } \: \: \: }$

━━━━━━━━━━━━━━━━

LEARN MORE FROM BRAINLY

If a : b=c:d=e: f= 2:1, then

(a + 2c + 3e) : (5b + 10d + 15f) =?

https://brainly.in/question/23293448