Question

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$\sf \footnotesize \bold{if \: α \: and \: β \: are \: the roots \: of \: the \: equation \:} \\ \sf \footnotesize \bold{ 375x ^{2} - 25x -2 = 0 then } \\ \sf \footnotesize \bold{ \: \displaystyle \lim_{\sf n\to \infty }\sum^{n}_{r=1}\alpha^r + \lim_{\sf n\to \infty... }\sum^{n}_{r=1}\beta^r} \\ \sf \footnotesize \bold{ \: is \: \: equal \: \: to}$


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Answer 1

ANSWER:

Given:

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

To Find:

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

Solution:

We are given that,

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

We know that,

⇒ α + β = - b/a = -(-25)/375 = 25/375 -----(1)

And,

⇒ αβ = c/a = -2/375 -----(2)

Now, we need to find the value of,

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

$\displaystyle\implies(\alpha+\alpha^2+\alpha^3+\dots+\infty)+(\beta+\beta^2+\beta^3+\dots+\infty)$

We can see that, each bracket is the sum of a Geometric Series till infinite terms.

So, for a Geometric Series: a, ar, ar², …… ∞,

$\displaystyle\hookrightarrow\sf{Sum}=\dfrac{a}{1-r}$

We had,

$\displaystyle\implies(\alpha+\alpha^2+\alpha^3+\dots+\infty)+(\beta+\beta^2+\beta^3+\dots+\infty)$

Here, in the first sum, a = α and r = α.

In the second sum, a = β and r = β.

So,

$\displaystyle\implies(\alpha+\alpha^2+\alpha^3+\dots\infty)+(\beta+\beta^2+\beta^3+\dots\infty)$

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

Taking LCM,

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

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

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

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

From (1) and (2),

$\displaystyle\implies\dfrac{\left(\dfrac{25}{375}\right)-2\left(\dfrac{-2}{375}\right)}{1-\left(\dfrac{25}{375}\right)+\left(\dfrac{-2}{375}\right)}$

$\displaystyle\implies\dfrac{\dfrac{25}{375}+\dfrac{4}{375}}{1-\dfrac{25}{375}-\dfrac{2}{375}}$

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

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

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

On dividing the fraction by 29,

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

Hence,

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

Answer 2

Step-by-step explanation:

1. Given a and Bare roots of quadratic equation

375x ^ 2 - 25x - 2 = 0 25 α+β 3755 and = alpha*beta = - 2/375 = 375 Now, 15

limn→∞ Σr_1 a² + limn→∞ Σr=1 ßr = (a + a² + a³ + + upto infinite terms) +

(a + a² + a³ + ... + upto infinite terms) +

= 1 1-8 : So = 1 for GP] a(1-3)+3(1-a) α-αβ+β-αβ (1-a)(1-3) (a+B)-2aß 1-(a+ß)+aß = 1-a-B+aß On substituting the value alpha + beta = 1/15 and alpha*beta = - 2/375 from Eqs. (i) and (ii) respectively, Eqs. =

(i) and (ii) respectively, 1-4 +375-25-2 15 = csec² x 2 X -Sec² 1 = 29 348 1 12 -

1/12