Question

Let \[\alpha \] and \[\beta \] are the roots of equation \[{{x}^{2}}-x-1=0\] with \[\alpha >\beta \], for all positive integers ‘n’ define \[{{a}_{n}}=\dfrac{{{\alpha }^{n}}-{{\beta }^{n}}}{\alpha -\beta },n\ge 1\], \[{{b}_{1}}=1,{{b}_{n}}={{a}_{n}}={{a}_{n+1}}-{{a}_{n-1}},n\ge 2\]. Then which of the following options is/are correct
(a) \[\sum\limits_{n=1}^{\infty }{\dfrac{{{a}_{n}}}{{{10}^{n}}}}=\dfrac{10}{89}\]
(b) \[{{b}_{n}}={{\alpha }^{n}}+{{\beta }^{n}},n\ge 1\]
(c) \[{{a}_{1}}+{{a}_{2}}+{{a}_{3}}+..............+{{a}_{n}}={{a}_{n+2}}-1,n\ge 1\]
(d) \[\sum\limits_{n=1}^{\infty }{\dfrac{{{b}_{n}}}{{{10}^{n}}}}=\dfrac{8}{89}\]

Answer 1

Hint: For solving this problem we will find the sum of roots \[\alpha +\beta \]and product of roots \[\alpha .\beta \] from the given equation. Then we use the given conditions in the question and try to check which of the options are correct. For finding the sum and product of roots we assume that the given equation is in the form of general quadratic equation \[a{{x}^{2}}+bx+c=0\] and use \[\alpha +\beta =\dfrac{-b}{a}\]and\[\alpha .\beta =\dfrac{c}{a}\], we solve the options taking one by one. For options (a) and (b) we use geometric progression (G,P) and use sum of G.P. that is \[a+ar+a{{r}^{2}}+..........+a{{r}^{n}}=\dfrac{a\left( {{r}^{n+1}}-1 \right)}{r-1}\] and if
\[a+ar+a{{r}^{2}}+..............\left( upto\text{ }\infty \right)=\dfrac{a}{1-r}if\left| r \right|\le 1\].

Complete step-by-step solution
Let us assume that the given equation \[{{x}^{2}}-x-1=0\] in the general form \[a{{x}^{2}}+bx+c=0\]
Then we can write sum of roots as
\[\begin{align}
  & \Rightarrow \alpha +\beta =\dfrac{-b}{a} \\
 & \Rightarrow \alpha +\beta =\dfrac{-\left( -1 \right)}{1}=1 \\
\end{align}\]
By taking the product of roots we will get
 \[\begin{align}
  & \Rightarrow \alpha .\beta =\dfrac{c}{a} \\
 & \Rightarrow \alpha .\beta =\dfrac{-1}{1}=-1 \\
\end{align}\]
Now we can take the LHS of each option and try to solve to prove that option.
Let us solve for option (a)
By taking option (a) and substituting \[{{a}_{n}}=\dfrac{{{\alpha }^{n}}-{{\beta }^{n}}}{\alpha -\beta },n\ge 1\] we will get
\[\begin{align}
  & \Rightarrow LHS=\sum\limits_{n=1}^{\infty }{\dfrac{{{a}_{n}}}{{{10}^{n}}}} \\
 & \Rightarrow LHS=\sum\limits_{n=1}^{\infty }{\dfrac{\left( \dfrac{{{\alpha }^{n}}-{{\beta }^{n}}}{\alpha -\beta } \right)}{{{10}^{n}}}} \\
\end{align}\]
Since \[\alpha -\beta \] is constant, by taking it outside the summation and separating the terms we will get
\[\Rightarrow LHS=\dfrac{1}{\alpha -\beta }\left( \sum\limits_{n=1}^{\infty }{{{\left( \dfrac{\alpha }{10} \right)}^{n}}}-\sum\limits_{n=1}^{\infty }{{{\left( \dfrac{\beta }{10} \right)}^{n}}} \right)\]
We know that \[\dfrac{\alpha }{10}\le \dfrac{\beta }{10}\le 1\] by using infinite GP formula mentioned above \[\begin{align}
  & \Rightarrow LHS=\dfrac{1}{\alpha -\beta }\left( \dfrac{\dfrac{\alpha }{10}}{1-\dfrac{\alpha }{10}}-\dfrac{\dfrac{\beta }{10}}{1-\dfrac{\beta }{10}} \right) \\
 & \Rightarrow LHS=\dfrac{1}{\alpha -\beta }\left( \dfrac{\alpha }{10-\alpha }-\dfrac{\beta }{10-\beta } \right) \\
 & \Rightarrow LHS=\dfrac{1}{\alpha -\beta }\left( \dfrac{-\alpha \beta +10\alpha +\alpha \beta -10\beta }{\alpha \beta -10\left( \alpha +\beta \right)+100} \right) \\
 & \Rightarrow LHS=\dfrac{1}{\alpha -\beta }\left( \dfrac{10\left( \alpha -\beta \right)}{\alpha \beta -10\left( \alpha +\beta \right)+100} \right) \\
\end{align}\]
By substituting the values of \[\alpha +\beta \] and \[\alpha .\beta \] in above equation we will get
\[\begin{align}
  & \Rightarrow LHS=\dfrac{10}{-1-10\left( 1 \right)+100} \\
 & \Rightarrow LHS=\dfrac{10}{89} \\
\end{align}\]
Therefore we can say that option (a) is correct.
We can say that option (b) is wrong because \[{{b}_{n}}=\dfrac{{{\alpha }^{n}}-{{\beta }^{n}}}{\alpha -\beta },n\ge 1\] and can never be \[{{b}_{n}}={{\alpha }^{n}}+{{\beta }^{n}},n\ge 1\].
By taking LHS of option (c) we will get
\[\begin{align}
  & \Rightarrow LHS={{a}_{1}}+{{a}_{2}}+{{a}_{3}}+..............+{{a}_{n}} \\
 & \Rightarrow LHS=\dfrac{\alpha -\beta }{\alpha -\beta }+\dfrac{{{\alpha }^{2}}-{{\beta }^{2}}}{\alpha -\beta }+...............+\dfrac{{{\alpha }^{n}}-{{\beta }^{n}}}{\alpha -\beta } \\
\end{align}\]
Taking \[\alpha -\beta \] out and separating the terms of \[\alpha \] and \[\beta \] we will get
\[\begin{align}
  & \Rightarrow LHS=\dfrac{1}{\alpha -\beta }\left( \left( \alpha +{{\alpha }^{2}}+{{\alpha }^{3}}+........+{{\alpha }^{n}} \right)-\left( \beta +{{\beta }^{2}}+{{\beta }^{3}}+........+{{\beta }^{n}} \right) \right) \\
 & \Rightarrow LHS=\dfrac{1}{\alpha -\beta }\left( \dfrac{\alpha \left( {{\alpha }^{n}}-1 \right)}{\alpha -1}-\dfrac{\beta \left( {{\beta }^{n}}-1 \right)}{\beta -1} \right) \\
 & \Rightarrow LHS=\dfrac{1}{\alpha -\beta }\left( \dfrac{\left( {{\alpha }^{n+1}}-\alpha \right)}{\alpha -1}-\dfrac{\left( {{\beta }^{n+1}}-\beta \right)}{\beta -1} \right) \\
\end{align}\]
By cross multiplying the terms inside the brackets we will get
\[\Rightarrow LHS=\dfrac{1}{\alpha -\beta }\left( \dfrac{\beta {{\alpha }^{n+1}}-\alpha \beta -{{\alpha }^{n+1}}+\alpha -\alpha {{\beta }^{n+1}}+\alpha \beta +{{\beta }^{n+1}}-\beta }{\alpha \beta -\left( \alpha +\beta \right)+1} \right)\]
By rearranging the terms in numerator and substituting \[\alpha +\beta \] and \[\alpha .\beta \] in above equation we will get
\[\begin{align}
  & \Rightarrow LHS=\dfrac{1}{\alpha -\beta }\left( \dfrac{\alpha \beta \left( {{\alpha }^{n}}-{{\beta }^{n}} \right)-\left( {{\alpha }^{n+1}}-{{\beta }^{n+1}} \right)+\left( \alpha -\beta \right)}{-1-\left( 1 \right)+1} \right) \\
 & \Rightarrow LHS=\dfrac{\left( {{\alpha }^{n}}-{{\beta }^{n}} \right)}{\alpha -\beta }+\dfrac{\left( {{\alpha }^{n+1}}-{{\beta }^{n+1}} \right)}{\alpha -\beta }-1 \\
 & \Rightarrow LHS={{a}_{n}}+{{a}_{n+1}}-1 \\
\end{align}\]
In the question we are given that \[{{a}_{n}}={{a}_{n+1}}-{{a}_{n-1}}\], replace ‘n’ by ‘n+1’ we get\[{{a}_{n+1}}={{a}_{n+2}}-{{a}_{n}}\].
By substituting this in above equation we will get
\[\begin{align}
  & \Rightarrow LHS={{a}_{n}}+{{a}_{n+2}}-{{a}_{n}}-1 \\
 & \Rightarrow LHS={{a}_{n+2}}-1 \\
\end{align}\]
Therefore option (c) is correct
We know that\[{{b}_{n}}={{a}_{n}}\].
As \[\sum\limits_{n=1}^{\infty }{\dfrac{{{a}_{n}}}{{{10}^{n}}}}=\dfrac{10}{89}\]then, \[\sum\limits_{n=1}^{\infty }{\dfrac{{{b}_{n}}}{{{10}^{n}}}}\ne \dfrac{8}{89}\]. So, option (d) is wrong.
Therefore option (a) and option (c) are correct answers.

Note: Some students will make mistakes in the solving part. Since the equations are very large and calculations in each option are very big we need to be careful in the calculations part. Also, in the formulae of infinite GP instead of taking \[\dfrac{a}{1-r}\] due to confusion students will take \[\dfrac{a}{r-1}\]. This part has to be taken care of in solving the question.