Question

If α and β are the roots of the equation 375x² – 25x – 2 = 0, then
limₙ→[infinity] ⁿ∑ᵣ₌₁ αʳ + limₙ→[infinity] ⁿ∑ᵣ₌₁ βʳ is equal to
(A) 1/12
(B) 29/358
(C) 7/116
(D) 21/346

Answer 1

Hey dude your answer is

a. 1/2

Hope this helps ❤️

Mark as brainliest ❤️

Answer 2

Option (A) is correct.

Step-by-step explanation:

Given that α and β are the roots of the equation 375x² – 25x – 2 = 0. So,

$\alpha +\beta =-\frac{-25}{375}=\frac{1}{15}$  and $\alpha \beta =\frac{-2}{375}$

Now,

$\lim_{n\rightarrow \infty }\sum_{r=1}^{n}\alpha ^{r}+\lim_{n\rightarrow \infty }\sum_{r=1}^{n}\beta ^{r}$

$=\left (\alpha+\alpha +\alpha ^{2}+.... \right )+\left ( \beta+\beta +\beta ^{2}+.... \right )$

The above series are the infinite geometric series, so

$\lim_{n\rightarrow \infty }\sum_{r=1}^{n}\alpha ^{r}+\lim_{n\rightarrow \infty }\sum_{r=1}^{n}\beta ^{r}\\=\frac{\alpha }{1-\alpha }+\frac{\beta }{1-\beta }$

$=\frac{\alpha }{1-\alpha }+\frac{\beta }{1-\beta }\\=\frac{\alpha -\alpha \beta +\beta -\beta }{1-\left ( \alpha +\beta \right )+\alpha \beta}$

$=\frac{\alpha +\beta -2\alpha \beta }{1-\left ( \alpha +\beta \right )+\alpha \beta }$

$=\frac{\frac{1}{15}-2\times \frac{-2}{375}}{1-\frac{1}{15}+\frac{-2}{375}}$

$=\frac{1}{12}$

Hence, option (A) is correct.