Question

if α and β are the roots of the equation 375x² - 25x -2 = 0 then
$\displaystyle \lim_{\sf n\to \infty }\sum^{n}_{r=1}\alpha^r + \lim_{\sf n\to \infty }\sum^{n}_{r=1}\beta^r$
is equal to

Answer 1

The roots of the equation $375x^2-25x-2=0$ are,

$\longrightarrow x=\dfrac{25\pm\sqrt{(-25)^2-(4\times375\times(-2))}}{2\times375}$

$\longrightarrow x=\dfrac{25\pm\sqrt{3625}}{750}$

$\longrightarrow x=\dfrac{5\pm\sqrt{145}}{150}$

So,

• $\alpha=\dfrac{5+\sqrt{145}}{150}$

• $\beta=\dfrac{5-\sqrt{145}}{150}$

We see $|\alpha|<1,\ |\beta|<1$ also.

Then,

$\displaystyle\longrightarrow\lim_{n\to\infty}\sum_{r=1}^n\alpha^r=\lim_{n\to\infty}\sum_{r=1}^n\left(\dfrac{5+\sqrt{145}}{150}\right)^r$

The sum actually represents an infinite geometric series of first term $\alpha$ and common ratio also $\alpha.$

We have the formula for infinite geometric series, that for $|k|<1,$

$\displaystyle\longrightarrow\lim_{n\to\infty}\sum_{r=1}^nak^{r-1}=\dfrac{a}{1-k}$

Here $a$ is first term and $k$ is common ratio.

If $a=k$ then,

$\displaystyle\longrightarrow\lim_{n\to\infty}\sum_{r=1}^nk^r=\dfrac{k}{1-k}$

Thus,

$\displaystyle\longrightarrow\lim_{n\to\infty}\sum_{r=1}^n\alpha^r=\dfrac{\frac{5+\sqrt{145}}{150}}{1-\frac{5+\sqrt{145}}{150}}$

$\displaystyle\longrightarrow\lim_{n\to\infty}\sum_{r=1}^n\alpha^r=\dfrac{5+\sqrt{145}}{145-\sqrt{145}}$

Similarly,

$\displaystyle\longrightarrow\lim_{n\to\infty}\sum_{r=1}^n\beta^r=\lim_{n\to\infty}\sum_{r=1}^n\left(\dfrac{5-\sqrt{145}}{150}\right)^r$

$\displaystyle\longrightarrow\lim_{n\to\infty}\sum_{r=1}^n\beta^r=\dfrac{\frac{5-\sqrt{145}}{150}}{1-\frac{5-\sqrt{145}}{150}}$

$\displaystyle\longrightarrow\lim_{n\to\infty}\sum_{r=1}^n\beta^r=\dfrac{5-\sqrt{145}}{145+\sqrt{145}}$

Now,

$\displaystyle\longrightarrow\lim_{n\to\infty}\sum_{r=1}^n\alpha^r+\lim_{n\to\infty}\sum_{r=1}^n\beta^r=\dfrac{5+\sqrt{145}}{145-\sqrt{145}}+\dfrac{5-\sqrt{145}}{145+\sqrt{145}}$

$\displaystyle\longrightarrow\lim_{n\to\infty}\sum_{r=1}^n\alpha^r+\lim_{n\to\infty}\sum_{r=1}^n\beta^r=\dfrac{2(145\times5+145)}{145^2-145}$

$\displaystyle\longrightarrow\lim_{n\to\infty}\sum_{r=1}^n\alpha^r+\lim_{n\to\infty}\sum_{r=1}^n\beta^r=\dfrac{2\times145\times6}{145\times144}$

$\displaystyle\longrightarrow\underline{\underline{\lim_{n\to\infty}\sum_{r=1}^n\alpha^r+\lim_{n\to\infty}\sum_{r=1}^n\beta^r=\dfrac{1}{12}}}$

Hence 1/12 is the answer.

Answer 2

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1/12