Question

if alpha and beta are the roots of the equation x^2+x-1=0, then find the value of ​

Answer 1

Answer:

The correct answer is option (b) .

Step-by-step explanation:

Given $\alpha,\beta$ are the roots of $x^2+x-1=0$

     $\Rightarrow \alpha+\beta=-1,\alpha\beta=-1$

Therefore

         $\displaystyle \lim_{n\to \infty}\sum_{r=1}^n(\alpha^r+\beta^r)$

       $=(\alpha+\beta)+(\alpha^2+\beta^2)+(\alpha^3+\beta^3)+...\infty\\=(\alpha+\alpha^2+\alpha^3+...)+(\beta+\beta^2+\beta^3+....)\\\\=\dfrac{\alpha}{1-\alpha}+\dfrac{\beta}{1-\beta}=\dfrac{\alpha(1-\beta)+\beta(1-\alpha)}{(1-\alpha)(1-\beta)}\\\\=\dfrac{(\alpha+\beta)-2\alpha\beta}{1-(\alpha+\beta)+\alpha\beta}$

Now substitute the values

       $=\dfrac{-1-2(-1)}{1-(-1)-1}=\dfrac{-1+2}{1+1-1}=1$

Answer 2

Given Question :-

$\rm \: \alpha , \beta \: are \: the \: roots \: of \: {x}^{2} + x - 1 = 0 \: then \: the \: value \: of \:$

$\rm \: \displaystyle\lim_{n \to \infty }\sum_{r=1}^n\rm ( { \alpha }^{r} + { \beta }^{r}) \: is \:$

(a) 0

(b) 1

(c) 2

(d) 3

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: \alpha , \beta \: are \: the \: roots \: of \: {x}^{2} + x - 1 = 0$

We know,

$\boxed{{\sf Sum\ of\ the\ roots=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

and

$\boxed{{\sf Product\ of\ the\ roots=\frac{Constant}{coefficient\ of\ x^{2}}}}$

So, using these results, we get

$\rm\implies \: \alpha + \beta = \dfrac{ - 1}{1} = - 1$

and

$\rm\implies \: \alpha \beta = \dfrac{ - 1}{1} = - 1$

Now, Consider

$\rm \: \displaystyle\lim_{n \to \infty }\sum_{r=1}^n\rm ( { \alpha }^{r} + { \beta }^{r})$

can be rewritten as

$\rm \: =  \: ( \alpha + \beta ) + ( { \alpha }^{2} + { \beta }^{2}) + ( { \alpha }^{3} + { \beta }^{3}) + - - - \infty$

$\rm \: =  \: ( \alpha + { \alpha }^{2} + { \alpha }^{3} + - - - \infty ) + ( \beta + { \beta }^{2} + { \beta }^{3} + - - - \infty )$

It forms an infinite GP series.

So, using sum of infinite GP series, we get

$\rm \: =  \: \dfrac{ \alpha }{1 - \alpha } + \dfrac{ \beta }{1 - \beta }$

$\rm \: =  \: \dfrac{ \alpha(1 - \beta ) + \beta (1 - \alpha ) }{(1 - \alpha)(1 - \beta ) }$

$\rm \: =  \: \dfrac{ \alpha - \alpha \beta + \beta - \alpha \beta }{1 - \alpha - \beta + \alpha \beta }$

$\rm \: =  \: \dfrac{ \alpha + \beta - 2 \alpha \beta }{1 - ( \alpha + \beta) + \alpha \beta }$

$\rm \: = \: \dfrac{ - 1 - 2( - 1)}{1 - ( - 1) - 1}$

$\rm \: = \: \dfrac{ - 1 + 2}{1}$

$\rm \: = \:1$

Hence,

$\rm\implies \: \boxed{\sf{  \:\: \: \rm \: \displaystyle\lim_{n \to \infty }\sum_{r=1}^n\rm ( { \alpha }^{r} + { \beta }^{r}) = 1 \: \: \: }}$

Thus, option (b) is correct.

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Formulae Used :-

Sum of infinite GP series having first term a and common ratio r, is given by

$\boxed{\sf{  \:\rm \: S_ \infty = \frac{a}{1 - r}, \: provided \: that \: |r| < 1 \: }}$