Question

If \[\alpha \] and \[\beta \] are the roots of the equation \[375{{x}^{2}}-25x-2=0\], then determine the value of \[\displaystyle \lim_{n \to \infty}\left( \sum\limits_{r=1}^{n}{{{\alpha }^{r}}} \right)+\displaystyle \lim_{n \to \infty}\left( \sum\limits_{r=1}^{n}{{{\beta }^{r}}} \right)\].
(a) \[\dfrac{21}{346}\]
(b) \[\dfrac{29}{358}\]
(c) \[\dfrac{1}{12}\]
(d) \[\dfrac{7}{116}\]

Answer 1

Hint: In this question, we are that \[\alpha \] and \[\beta \] are the roots of the equation \[375{{x}^{2}}-25x-2=0\]. Now for any quadratic equation \[a{{x}^{2}}+bx+c=0\], if \[p\] and \[q\] are the roots of the equation \[a{{x}^{2}}+bx+c=0\] then we have the sum of the roots \[p\] and \[q\] is given by \[p+q=-\dfrac{b}{a}\] and the products of the roots \[p\] and \[q\] is given by \[pq=\dfrac{c}{a}\]. We will also use the fact that the sum of an infinite geometric series \[\sum\limits_{k=1}^{\infty }{{{a}^{k}}}=\dfrac{a}{1-a}\] whenever the value of \[a\] is less than 1.

Complete step by step answer:
We are given the equation \[375{{x}^{2}}-25x-2=0\].
It is also given that \[\alpha \] and \[\beta \] are the roots of the equation \[375{{x}^{2}}-25x-2=0\].
Since we know that for any quadratic equation \[a{{x}^{2}}+bx+c=0\], if \[p\] and \[q\] are the roots of the equation \[a{{x}^{2}}+bx+c=0\] then we have the sum of the roots \[p\] and \[q\] is given by \[p+q=-\dfrac{b}{a}\] and the products of the roots \[p\] and \[q\] is given by \[pq=\dfrac{c}{a}\].
Now comparing quadratic equation \[a{{x}^{2}}+bx+c=0\] with the given equation \[375{{x}^{2}}-25x-2=0\], we will have
\[a=375,b=-25\] and \[c=-2\].
Now since \[\alpha \] and \[\beta \] are the roots of the equation \[375{{x}^{2}}-25x-2=0\].
Therefore we have
\[\begin{align}
  & \alpha +\beta =-\dfrac{b}{a} \\
 & =\dfrac{25}{375}
\end{align}\]
And also
The product of the roots \[\alpha \] and \[\beta \] is given by
\[\begin{align}
  & \alpha \beta =\dfrac{c}{a} \\
 & =-\dfrac{2}{375}
\end{align}\]
Now consider the expression \[\displaystyle \lim_{n \to \infty}\left( \sum\limits_{r=1}^{n}{{{\alpha }^{r}}} \right)+\displaystyle \lim_{n \to \infty}\left( \sum\limits_{r=1}^{n}{{{\beta }^{r}}} \right)\].
Since we know that \[\displaystyle \lim_{n \to \infty}\left( \sum\limits_{r=1}^{n}{{{\alpha }^{r}}} \right)=\sum\limits_{r=1}^{\infty }{{{\alpha }^{r}}}\] and \[\displaystyle \lim_{n \to \infty}\left( \sum\limits_{r=1}^{n}{{{\beta }^{r}}} \right)=\sum\limits_{r=1}^{\infty }{{{\beta }^{r}}}\], therefore we have
\[\displaystyle \lim_{n \to \infty}\left( \sum\limits_{r=1}^{n}{{{\alpha }^{r}}} \right)+\displaystyle \lim_{n \to \infty}\left( \sum\limits_{r=1}^{n}{{{\beta }^{r}}} \right)=\sum\limits_{r=1}^{\infty }{{{\alpha }^{r}}}+\sum\limits_{r=1}^{\infty }{{{\beta }^{r}}}.........(1)\]
Also since we know that sum of an infinite geometric series \[\sum\limits_{k=1}^{\infty }{{{a}^{k}}}=\dfrac{a}{1-a}\] whenever the value of \[a\] is less than 1.
Here we have the geometric series \[\sum\limits_{r=1}^{\infty }{{{\alpha }^{r}}}\] and \[\sum\limits_{r=1}^{\infty }{{{\beta }^{r}}}\] with common ration \[\alpha <1\] and \[\beta <1\] respectively.
Therefore the sum \[\sum\limits_{r=1}^{\infty }{{{\alpha }^{r}}}=\dfrac{\alpha }{1-\alpha }.......(2)\]
And \[\sum\limits_{r=1}^{\infty }{{{\beta }^{r}}}=\dfrac{\beta }{1-\beta }.......(3)\]
Now on substituting the values of equation (2) and equation (3) in equation (1), we get
\[\displaystyle \lim_{n \to \infty}\sum\limits_{r=1}^{n}{{{\alpha }^{r}}}+\displaystyle \lim_{n \to \infty}\sum\limits_{r=1}^{n}{{{\beta }^{r}}}=\dfrac{\alpha }{1-\alpha }+\dfrac{\beta }{1-\beta }\]
On simplifying the above expression, we get
\[\begin{align}
  & \displaystyle \lim_{n \to \infty}\left( \sum\limits_{r=1}^{n}{{{\alpha }^{r}}} \right)+\displaystyle \lim_{n \to \infty}\left( \sum\limits_{r=1}^{n}{{{\beta }^{r}}} \right)=\dfrac{\alpha }{1-\alpha }+\dfrac{\beta }{1-\beta } \\
 & =\dfrac{\alpha \left( 1-\beta \right)+\beta \left( 1-\alpha \right)}{\left( 1-\alpha \right)\left( 1-\beta \right)} \\
 & =\dfrac{\alpha +\beta -2\alpha \beta }{1-\alpha -\beta +\alpha \beta } \\
 & =\dfrac{\alpha +\beta -2\alpha \beta }{1-\left( \alpha +\beta \right)+\alpha \beta }
\end{align}\]
Now on substituting the values \[\alpha +\beta =\dfrac{25}{375}\] and \[\alpha \beta =-\dfrac{2}{375}\] in the above equation we get
\[\begin{align}
  & \displaystyle \lim_{n \to \infty}\left( \sum\limits_{r=1}^{n}{{{\alpha }^{r}}} \right)+\displaystyle \lim_{n \to \infty}\left( \sum\limits_{r=1}^{n}{{{\beta }^{r}}} \right)=\dfrac{\alpha +\beta -2\alpha \beta }{1-\left( \alpha +\beta \right)+\alpha \beta } \\
 & =\dfrac{\dfrac{25}{375}-2\left( \dfrac{-2}{375} \right)}{1-\left( \dfrac{25}{375} \right)+\left( \dfrac{-2}{375} \right)}
\end{align}\]
On simplifying the above equation, we will have
\[\begin{align}
  & \displaystyle \lim_{n \to \infty}\left( \sum\limits_{r=1}^{n}{{{\alpha }^{r}}} \right)+\displaystyle \lim_{n \to \infty}\left( \sum\limits_{r=1}^{n}{{{\beta }^{r}}} \right)=\dfrac{25+4}{375-25-2} \\
 & =\dfrac{29}{348} \\
 & =\dfrac{1}{12}
\end{align}\]
Therefore we have that the value of \[\displaystyle \lim_{n \to \infty}\left( \sum\limits_{r=1}^{n}{{{\alpha }^{r}}} \right)+\displaystyle \lim_{n \to \infty}\left( \sum\limits_{r=1}^{n}{{{\beta }^{r}}} \right)\] is equals to \[\dfrac{1}{12}\].

So, the correct answer is “Option C”.

Note: In this problem, in order to determine the value of \[\displaystyle \lim_{n \to \infty}\left( \sum\limits_{r=1}^{n}{{{\alpha }^{r}}} \right)+\displaystyle \lim_{n \to \infty}\left( \sum\limits_{r=1}^{n}{{{\beta }^{r}}} \right)\] we have to first evaluate the limit to get \[\displaystyle \lim_{n \to \infty}\left( \sum\limits_{r=1}^{n}{{{\alpha }^{r}}} \right)+\displaystyle \lim_{n \to \infty}\left( \sum\limits_{r=1}^{n}{{{\beta }^{r}}} \right)=\sum\limits_{r=1}^{\infty }{{{\alpha }^{r}}}+\sum\limits_{r=1}^{\infty }{{{\beta }^{r}}}\] and then proceed using the fact that the sum of an infinite geometric series \[\sum\limits_{k=1}^{\infty }{{{a}^{k}}}=\dfrac{a}{1-a}\] whenever the value of \[a\] is less than 1.