Question

Let $\alpha$ and $\beta$ be the roots of $x^2–x–1=0$, with $\alpha >\beta$. For all positive integers $n$, define

$\begin{array}{l}{a}_n=\frac{{\alpha }^n–{\beta }^n}{\alpha –\beta },n\ge 1\end{array}$

$b_1=1$ and $b_n=a_{n-1}+a_{n+1},n\ge 2.$

Then which of the following options is/are correct?

• A. $\begin{array}{l}{a}_1+a_2+a_3+\dots .+a_n=a_{n+2}–1\end{array},n\ge 1$
• B. $\begin{array}{l}\sum _{n=1}^{\mathrm{\infty }}\frac{a_n}{{10}^n}=\frac{10}{89}\end{array}$
• C. $\begin{array}{l}\sum _{n=1}^{\mathrm{\infty }}\frac{b_n}{{10}^n}=\frac{8}{89}\end{array}$
• D. $\begin{array}{l}{b}_n={\alpha }^n+{\beta }^n\end{array},n\ge 1$

Answer 1

Answer: (A), (B), (D)

$\begin{array}{l}{b}_n={\alpha }^n+{\beta }^n\end{array},n\ge 1$

Determining the correct statements:

Option(A):

$$\begin{gathered} a_1+a_2+....+a_n=\sum a_i \\ =\frac{\sum {\alpha }_i-\sum {\beta }_i}{\alpha -\beta }\left[\begin{array}{l}\because a_n=\frac{{\alpha }^n–{\beta }^n}{\alpha –\beta },n\ge 1\end{array}\right] \end{gathered}$$

Since sum of $nth$ term of geometrical progression with first term$a$ and common ratio$r$ $=\frac{a(1-r^n)}{1-r},r<1$

$$\begin{gathered} \Rightarrow a_1+a_2+....+a_n=\frac{\frac{\alpha (1-{\alpha }^n)}{1-\alpha }-\frac{\beta (1-{\beta }^n)}{1-\beta }}{\alpha -\beta } \\ =\frac{(1+\alpha )(1-{\alpha }^n)-(\beta +1)(1-{\beta }^n)}{(1-\alpha )(1-\beta )(\alpha -\beta )} \\ =\frac{{\alpha }^2-{\alpha }^{n+2}-{\beta }^2+{\beta }^{n+2}}{(1-\alpha )(1-\beta )(\alpha -\beta )} \\ =\frac{{\alpha }^2-{\beta }^2+{\alpha }^{n+2}-{\beta }^{n+2}}{(1-\alpha )(1-\beta )(\alpha -\beta )} \end{gathered}$$

Since we know that difference of roots of $a{x}^2+bx+c=\frac{\sqrt{b^2-4ac}}{a}$ therefore $\alpha -\beta =\sqrt{5}$

$$\begin{gathered} \Rightarrow a_1+a_2+....+a_n=\frac{\sqrt{5}+{\alpha }^{n+2}-{\beta }^{n+2}}{\beta -\alpha } \\ =-1+a_{n+2} \\ \Rightarrow a_1+a_2+....+a_n=-1+a_{n+2} \end{gathered}$$

Hence, option (A) is correct.

Option(B):

$\sum _{n=1}^{\infty }\frac{a_n}{{10}^n}=\sum \frac{{\alpha }_n-{\beta }_n}{(\alpha -\beta ){10}^n}$

Since we know infinite sum of GP with term $a$ with common ration $r$ =$\frac{a}{1-r}$

$$\begin{gathered} \sum _{n=1}^{\infty }\frac{a_n}{{10}^n}=\frac{1}{\alpha -\beta }\left(\frac{{\displaystyle \frac{\alpha }{10}}}{1-{\displaystyle \frac{\alpha }{10}}}-\frac{{\displaystyle \frac{\beta }{10}}}{1-{\displaystyle \frac{\beta }{10}}}\right)\left[\because \alpha \text{and}\beta \text{are the roots}\right] \\ =\frac{1}{\alpha -\beta }\left(\frac{{\displaystyle \alpha }}{10-{\displaystyle \alpha }}-\frac{{\displaystyle \beta }}{10-{\displaystyle \beta }}\right) \\ =\frac{1}{\alpha -\beta }\left(\frac{{\displaystyle 10(\alpha -\beta )-\alpha \beta +\alpha \beta }}{100-10(\alpha +\beta )+\alpha \beta }\right) \\ =\frac{1}{\sqrt{5}}\left(\frac{{\displaystyle 10\sqrt{5}}}{100-10-1}\right)\left[\because \alpha +\beta =1,\alpha \beta =-1,\&\alpha -\beta =\sqrt{5}\right] \\ =\frac{10}{89} \end{gathered}$$

Hence, option (B) is correct.

Option(D):

$$\begin{gathered} b_n=a_{n+1}+a_{n-1} \\ =\frac{{\alpha }_{n+1}-{\beta }_{n+1}}{\alpha -\beta }+\frac{{\alpha }_{n-1}-{\beta }_{n-1}}{\alpha -\beta }\left[\begin{array}{l}\because a_n=\frac{{\alpha }^n–{\beta }^n}{\alpha –\beta },n\ge 1\end{array}\right] \\ =\frac{{\alpha }_{n-1}(\alpha +2)-{\beta }_{n-1}(\beta +2)}{\alpha -\beta } \\ =\frac{{\alpha }_{n-1}\left(\frac{5+\sqrt{5}}{2}\right)-{\beta }_{n-1}\left(\frac{5-\sqrt{5}}{2}\right)}{\alpha -\beta }\left[Bysolving\alpha +\beta =1\&\alpha -\beta =\sqrt{5}\right] \\ =\frac{\sqrt{5}{\alpha }_{n-1}\left(\frac{5+1}{2}\right)-\sqrt{5}{\beta }_{n-1}\left(\frac{5-1}{2}\right)}{\alpha -\beta } \\ =\frac{\sqrt{5}\left({\alpha }_n+{\beta }_n\right)}{\alpha -\beta } \\ {b}_n=\left({\alpha }_n+{\beta }_n\right) \end{gathered}$$

Hence, option (D) is correct .

Option(C):

$$\begin{gathered} \begin{array}{l}\sum _{n=1}^{\mathrm{\infty }}\frac{b_n}{{10}^n}=\end{array}\sum {\left(\frac{\alpha }{10}\right)}^n+\sum {\left(\frac{\beta }{10}\right)}^n[\text{from option}(D\left)\right] \\ =\frac{\frac{\alpha }{10}}{1-\frac{\alpha }{10}}+\frac{\frac{\beta }{10}}{1-\frac{\beta }{10}}[\text{sum of infinite}GP] \\ =\frac{\alpha }{1-\alpha }+\frac{\beta }{1-\beta } \\ =\frac{10(\alpha +\beta )-2\alpha \beta }{100-10(\alpha +\beta )+\alpha \beta } \\ =\frac{10+2}{89} \\ =\frac{12}{89}\ne \frac{8}{89} \end{gathered}$$

Thus, option (C) is incorrect .

Hence, options (A), (B), and (D) are correct.