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Algebra Level 3

If $\frac { x }{ a } + \frac { y }{ b } =1$ and $\frac { { x }^{ 2 } }{ a } + \frac { { y }^{ 2 } }{ b } = \frac { ab }{ a+b }$ then the value of $\frac { { x }^{ n+1 } }{ a } + \frac { { y }^{ n+1 } }{ b }$ is:

{( \frac { ab }{ a+b } )}^{ n }

{( \frac { a-b }{ ab } )}^{ n }

{( \frac { a+b }{ ab } )}^{ n }

{( \frac { ab }{ a-b } )}^{ n }

3 solutions

Pranjal Jain
Jan 18, 2015

As a 15 second solution, Dimension Analysis can work here, assume dimensions of x, y, a, b be [L]. Then the dimension of required expression is $\dfrac{[L]^{n+1}}{[L]}=[L]^n$ .

Now we are left with two options. There seems nothing wrong with $a=b$ , so only option left is $\left (\dfrac{ab}{a+b}\right)^n$

Ha ha , even I used dimensional analysis . Though I was hesitant to use it since I had only one shot at it , but yeah I took the chance and it paid off .

A Former Brilliant Member - 6 years, 4 months ago

Why hesitation? I mean it seems quite clear.

Pranjal Jain - 6 years, 4 months ago

I was hesitating because I had never used dimensional analysis in a Mathematics question before though I frequently use it in Physics . But yeah, now I'll be sure to use it as often as needed in any question .

A Former Brilliant Member - 6 years, 4 months ago

Congratulations. Nice way of thinking.

Niranjan Khanderia - 6 years, 4 months ago

Wow!!!!!! That's great. Nice solution...

Aditya Tiwari - 6 years, 4 months ago

Or any other way to solve this?? I just made a guess and find the answer !

Mahtab Hossain - 6 years, 1 month ago

Raghav Vaidyanathan
Jan 20, 2015

Put $x=y=0.5$ and $a=b=1$ . Only one of the options will match. The values are taken such that they conform to the condition given in the question.

How can you take x=y=0.5 and a=b =1 ? And from there how do you find
$\frac { { x }^{ n+1 } }{ a } + \frac { { y }^{ n+1 } }{ b }$ ?

Niranjan Khanderia - 6 years, 4 months ago

by putting values as above: $\frac { x }{ a } + \frac { y }{ b } =0.5/1+0.5/1=1$ and

$\frac { { x }^{ 2 } }{ a } + \frac { { y }^{ 2 } }{ b } =0.25+0.25=0.5= \frac { ab }{ a+b }$

$\frac { { x }^{ n+1 } }{ a } + \frac { { y }^{ n+1 } }{ b }= 2(0.5)^{n+1}=(0.5)^n={ (\frac { ab }{ a+b })}^n$

That is the only option which gives value $(0.5)^n$

Raghav Vaidyanathan - 6 years, 4 months ago

On what bases do you chose your values? If we assume x=y and a=b according to you, we get:-

$\therefore~ \dfrac { x }{ a } + \dfrac { y }{ b }=2*\dfrac { x }{ a }\\and ~ \dfrac { ab }{ a+b }=(\dfrac { a }{ 2})......(1)\\\therefore~\dfrac { x }{ a }=\dfrac { 1 }{2},~this ~satisfys~ both~ the~ given~ conditions.\\Also~\dfrac { x^{ n+1 } }{ a } + \dfrac { y^{ n+1 } }{ b} =2*\dfrac{ x }{ a }*x^n=x^n\\= ( \dfrac{1}{2})^n*a^n=(\dfrac{a}{2})^n=(\dfrac { ab }{ a+b })^n..by ~(1)\\\text{I do not see any need to give any values to x,y, and a,b.}\\\text{ Though ratio x:a::1:2 should be observed.}\\\text{ But there is no justification for assuming x=y and a=b. }$

Niranjan Khanderia - 6 years, 4 months ago

Niranjan Khanderia
Jan 18, 2015

$\dfrac { x }{ a } + \dfrac { y }{ b } =1~~~\implies\dfrac { x }{ a } + \dfrac { y } { b } = (\dfrac { ab }{ a+b })^{ \color{#D61F06}{0} }~~ and\\$ $\dfrac { x^2 }{ a } + \dfrac { y^2}{ b } =( \dfrac { ab }{ a+b })^{\color{#624F41}{1} } .\\~ then ~the ~value~ of\\\dfrac{ x^{ n+1 } }{ a } + \dfrac { y^{ n+1 } }{ b } =~~~~~\boxed{ ~~\huge{ (\dfrac { ab }{ a+b } )^{\color{#3D99F6}{n} } } }$

Sorry but this is no method, I mean its much like a guess.

Pranjal Jain - 6 years, 4 months ago

There is a logic as per my thinking. The logic is, "The expont for $\dfrac{ab}{a+b}$ is less than exponent of x and y by 1, $\text{while that of a and b at the denominator remains the same. }$ "

Niranjan Khanderia - 6 years, 4 months ago

Exactly! I got that! But using given equation in this way is no logic according to me. You should prove that this will always happen.

Pranjal Jain - 6 years, 4 months ago

Dare to do this in 15 seconds?

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