Question

If alpha and beta are the roots of x^2+x-1=0 then the value of limit n to Indy summation r = 1 to n (α^r+β^r) =​

Answer 1

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

Given that

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

So,

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

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

And

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

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

Now, Consider

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

$\rm \:  =\displaystyle\lim_{n \to \infty }\bigg[( \alpha + \beta) + ( { \alpha }^{2}+{ \beta }^{2}) + - - + ( { \alpha }^{n} + { \beta }^{n}) \bigg]$

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

Now, its an infinite GP series. So, using sum of infinite GP, 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,

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