JEE MAIN - Mathematics (2019 - 12th April Morning Slot - No. 2)
If $$\alpha $$ and $$\beta $$ are the roots of the equation 375x2
– 25x – 2 = 0, then $$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{\alpha ^r}} + \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{\beta ^r}} $$ is equal to :
$${7 \over {116}}$$
$${{29} \over {348}}$$
$${1 \over {12}}$$
$${{21} \over {346}}$$
Explanation
$$\alpha $$ and $$\beta $$ are the two root of 375x2 - 25x - 2 = 0
Both of the roots are lie in (-1, 1) hence sum of given series is finite
$$\mathop {\lim }\limits_{n \to \infty } \left( {{\alpha \over {1 - \alpha }} + {\beta \over {1 - \beta }}} \right) = {{\alpha (1 - \beta ) + \beta (1 - \alpha )} \over {(1 - \alpha )(1 - \beta )}}$$
$$ \Rightarrow {{\left( {\alpha + \beta } \right) - 2\alpha \beta } \over {1 - (\alpha + \beta ) + \alpha \beta }} = {{25 - 2( - 2)} \over {375 - 25 - 2}} = {{29} \over {348}}$$
Both of the roots are lie in (-1, 1) hence sum of given series is finite
$$\mathop {\lim }\limits_{n \to \infty } \left( {{\alpha \over {1 - \alpha }} + {\beta \over {1 - \beta }}} \right) = {{\alpha (1 - \beta ) + \beta (1 - \alpha )} \over {(1 - \alpha )(1 - \beta )}}$$
$$ \Rightarrow {{\left( {\alpha + \beta } \right) - 2\alpha \beta } \over {1 - (\alpha + \beta ) + \alpha \beta }} = {{25 - 2( - 2)} \over {375 - 25 - 2}} = {{29} \over {348}}$$
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