Question

If α and β are the roots of the equation 375x² - 25x - 2 = 0, then $\displaystyle{\rm{ \lim_{n \to \infty} \sum\limits_{r=1}^{n} \alpha^r+\lim_{n \to \infty} \sum\limits_{r=1}^{n} \beta^r }}$ is equal to? [JEE(Main)-Apr 2019]

Answer 1

◘ Given ◘

α and β are the roots of the equation 375x² - 25x - 2 = 0.

◘ To Find ◘

The value of :

$\displaystyle \sf{ \lim_{n \to \infty} \sum\limits_{r=1}^{n}\alpha^r+\lim_{n \to \infty} \sum\limits_{r=1}^{n}\beta^r }$

◘ Solution ◘

For the given equation 375x² - 25x - 2 = 0,

α + β = -b/a

⇒ α + β = -(-25)/375

⇒ α + β = 25/375

⇒ α + β = 1/15

_____________

αβ = c/a

⇒ αβ = -2/375

_________________

$\displaystyle \sf{ \lim_{n \to \infty} \sum\limits_{r=1}^{n}\alpha^r+\lim_{n \to \infty} \sum\limits_{r=1}^{n}\beta^r }$

$\sf{=(\alpha^1+\alpha^2+\alpha^3+ \dots)+(\beta^1+\beta^2+\beta^3+ \dots)}$

For sum of infinite series of a GP,

$\displaystyle {\sf{= \frac{\alpha}{1-\alpha}+\frac{\beta}{1-\beta} }}\\\\\displaystyle {\sf{= \frac{\alpha-\alpha \beta+\beta-\alpha\beta}{(1-\alpha)(1-\beta)} }}\\\\\displaystyle {\sf{= \frac{\alpha+\beta-2\alpha\beta}{1-\beta-\alpha+\alpha\beta} }}\\\\\displaystyle {\sf{= \frac{\alpha+\beta-2\alpha\beta}{1-(\alpha+\beta)+\alpha\beta} }}$

Putting the values from above :-

$\displaystyle {\sf{= \frac{\frac{1}{15} -2\times \frac{-2}{375} }{1-(\frac{1}{15} )+(\frac{-2}{375}) } }}\\\\\\\displaystyle {\sf{= \frac{\frac{1}{15}+\frac{4}{375} }{\frac{375-25-2}{375} } }}\\\\\\\displaystyle {\sf{= \frac{\frac{25+4}{375} }{\frac{348}{375} } }}\\\\\\\displaystyle {\sf{= \frac{29}{375} \div \frac{348}{375} }}\\\\\displaystyle {\sf{= \frac{29}{375} \times \frac{375}{348} }}\\\\\displaystyle {\sf{= \frac{29}{348} }}$

$\boxed{\boxed{\displaystyle {\sf{= \frac{1}{12} }}}}$

Hence, the answer is 1/12.

Answer 2

In the above Question , the following information is given -

α and β are the roots of the equation 375x² - 25x - 2 = 0

To find -

Find the value of -

$\displaystyle{\rm{ \lim_{n \to \infty} \sum\limits_{r=1}^{n} \alpha^r+\lim_{n \to \infty} \sum\limits_{r=1}^{n} \beta^r }}$

Solution -

Given quadratic equation -

375x² - 25x - 2 = 0

Using Vieta ' s relations -

Sum of Zeroes -

=> α + β

=> ( - b / a )

=> ( 25 / 375 )

So ,

$\sf{ \alpha + \beta = \dfrac{ 25 }{ 375 } }$

Similarly -

Product of Zeroes -

=> α β

=> ( c / a )

=> ( -2 / 375 )

So ,

$\sf{ \alpha \times \beta = \dfrac{ -2 }{ 375 } }$

Now , we have been asked to find the value of -

$\displaystyle{\rm{ \lim_{n \to \infty} \sum\limits_{r=1}^{n} \alpha^r+\lim_{n \to \infty} \sum\limits_{r=1}^{n} \beta^r }}$

Now ,

Here , this can be seen as a case of an infinite gp series .

Sum of the term a -

=> $\sf{ \dfrac{ \alpha }{ 1 - \alpha } }$

Sum of the term β -

=> $\sf{ \dfrac{ \beta }{ 1 - \beta } }$

Total sum -

$\sf{ \dfrac{ \alpha }{ 1 - \alpha } + \dfrac{ \beta }{ 1 - \beta } } \\ \\ \sf{ \implies { \dfrac{ \alpha + \beta - 2 \alpha \beta } { 1 - \alpha - \beta + \alpha \beta } }} \\ \\ \sf{ \implies { \dfrac{ 1 }{ 12 } }}$ [/tex]