$\displaystyle{-2)-2)-2)-2}$

Algebra Level 5

The coefficient of $\displaystyle{x^2}$ appearing after parenthesis have been removed and like terms have been collected in $\underbrace{\Big( ..............(x-2)^{2} -2)^{2} - 2)^{2}..)......\Big)^{2} }_\text{k times}$ can be expressed as $\dfrac{\alpha^{2k-1} - \alpha^{k-1}}{\beta}$ . Where $\displaystyle{\alpha}$ and $\displaystyle{\beta}$ are positive coprime integers. Give their sum.

The answer is 7.

1 solution

Aman Sharma
Nov 24, 2014

I know it is not a correct explaination but this is how i solved this question:-

First of all note that question says that the coefficient of $x^2$ can be expressed as $\frac{\alpha^{2k-1}-\alpha^{k-1}}{\beta}............(1)$ Now for $k=1,2$ coefficient of $x^2$ is 1 and 20 respectivaly.Now putting these values in equation one we can form a system of two equation with two variable:- $\frac{\alpha-1}{\beta}=1.....(2)$ $\frac{\alpha^3-\alpha}{\beta}=20....(3)$ Solving above system of equations only..working values of $\alpha$ and $\beta$ i found was 4 and 3

@Abhishek Singh it is my humble request please post its solution so that beginers like me can learn from you

Did the same thing

prajwal kavad - 6 years, 5 months ago

I realy want to know how we actualy solve such questions

Aman Sharma - 6 years, 5 months ago

− 2 ) − 2 ) − 2 ) − 2 \displaystyle{-2)-2)-2)-2} − 2 ) − 2 ) − 2 ) − 2

The answer is 7.

1 solution

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$\displaystyle{-2)-2)-2)-2}$