Question

Q46. If $\alpha$ and $\beta$ are the roots of the equation 375x² – 25x – 2 = 0, then $\displaystyle{\lim_{n\to\infty}\sum_{r=1}^n\alpha^r+\lim_{n\to\infty}\sum_{r=1}^n\beta^r}$ is equal to:

Ⓐ 7/116
Ⓑ 29/348
Ⓒ 1/12
Ⓓ 21/346​

Answer 1

ANSWER:

Given:

• $\alpha$ and $\beta$ are zeroes of 375x² – 25x – 2 = 0

To Find:

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

Solution:

We are given that, $\alpha$ and $\beta$ are zeroes of 375x² – 25x – 2 = 0.

We can have,

$\implies\text{Sum of zeroes}=\alpha+\beta=\dfrac{\text{-coefficient of x}}{\text{coefficient of x ^2 }}$

So,

$\implies\alpha+\beta=\dfrac{-(-25)}{375}=\dfrac{25}{375}$

$\implies\alpha+\beta=\dfrac{1}{15}- - - -(1)$

And,

$\implies\text{Product of zeroes}=\alpha\beta=\dfrac{\text{constant}}{\text{coefficient of x ^2 }}$

$\implies\alpha\beta=\dfrac{-2}{375}- - - -(2)$

Now, we need to find value of,

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

We can re-write it as,

$\displaystyle{\longrightarrow\sum_{r=1}^{\infty}\alpha^r+\sum_{r=1}^{\infty}\beta^r}$

Now, we open the summations,

$\displaystyle{\implies\bigg(\alpha^1+\alpha^2+\alpha^3+\dots+\alpha^{\infty}\bigg)+\bigg(\beta^1+\beta^2+\beta^3+\dots+\beta^{\infty}\bigg)}$

We can see that, terms in each summation forms a GP, with $\alpha$ and $\beta$ as the common ratios respectively.

We know that,

$\hookrightarrow\text{Sum of infinite terms of GP}=\dfrac{a}{1-r}$

Here, a is first term and r is common ratio.

Applying this,

$\displaystyle{\implies\bigg(\alpha^1+\alpha^2+\alpha^3+\dots+\alpha^{\infty}\bigg)+\bigg(\beta^1+\beta^2+\beta^3+\dots+\beta^{\infty}\bigg)}$

$\implies\bigg(\dfrac{\alpha}{1-\alpha}\bigg)+\bigg(\dfrac{\beta}{1-\beta}\bigg)$

So,

$\displaystyle\implies\dfrac{\alpha}{1-\alpha}+\dfrac{\beta}{1-\beta}$

Taking LCM,

$\displaystyle \implies\dfrac{(\alpha)(1-\beta)+(\beta)(1-\alpha)}{(1-\alpha)(1-\beta)}$

$\displaystyle \implies\dfrac{\alpha-\alpha\beta+\beta-\alpha\beta}{1-\alpha-\beta+\alpha\beta}$

Regrouping,

$\displaystyle \implies\dfrac{(\alpha+\beta)-2(\alpha\beta)}{1-(\alpha+\beta)+(\alpha\beta)}$

Substituting values from (1) & (2),

$\displaystyle \implies\dfrac{\left(\frac{1}{15}\right)-2\left(\frac{-2}{375}\right)}{1-\left(\frac{1}{15}\right)+\left(\frac{-2}{375}\right)}$

$\displaystyle \implies\dfrac{\frac{1}{15}+\frac{4}{375}}{1-\frac{1}{15}-\frac{2}{375}}$

Taking LCM,

$\displaystyle \implies\dfrac{\frac{25+4}{375}}{\frac{375-25-2}{375}}$

$\displaystyle \implies\dfrac{\frac{29}{375}}{\frac{348}{375}}$

On cancelling 375 in both numerator and denominator,

$\implies\dfrac{29}{348}$

On simplifying,

$\implies\bf\dfrac{1}{12}$

Therefore,

$\displaystyle{\bf\implies\lim_{n\to\infty}\sum_{r=1}^n\alpha^r+\lim_{n\to\infty}\sum_{r=1}^n\beta^r\:\:\:=\:\:\:\dfrac{1}{12}}$

Answer 2

Answer:

see Image for answer

please follow