Question

Suppose that \[F\left( {n + 1} \right) = \dfrac{{2F\left( n \right) + 1}}{2}\] for \[n = 1,2,3, \ldots \] and \[F\left( 1 \right) = 2\]. Then, \[F\left( {101} \right)\] equals
(a) 50
(b) 52
(c) 54
(d) None of these

Answer 1

Hint: Here, we have to find the value of the expression \[F\left( {101} \right)\]. We will use the given equation to find the value of the function at the values of \[n\] as 1, 2, 3, 4. Then, we will use this information to form an A.P. Finally, we will use the formula for \[{n^{{\rm{th}}}}\] term of an A.P. to find the value of the expression \[F\left( {101} \right)\].

Formula Used:
We will use the formula of the \[{n^{{\rm{th}}}}\] term of an A.P. is given by the formula \[{a_n} = a + \left( {n - 1} \right)d\], where \[a\] is the first term of the A.P. and \[d\] is the common difference.

Complete step-by-step answer:
It is given that \[n = 1,2,3, \ldots \].
Substituting \[n = 1\] in the equation \[F\left( {n + 1} \right) = \dfrac{{2F\left( n \right) + 1}}{2}\], we get
\[ \Rightarrow F\left( {1 + 1} \right) = \dfrac{{2F\left( 1 \right) + 1}}{2}\]
\[ \Rightarrow F\left( 2 \right) = \dfrac{{2F\left( 1 \right) + 1}}{2}\]
Substituting \[F\left( 1 \right) = 2\] in the expression, we get
\[ \Rightarrow F\left( 2 \right) = \dfrac{{2 \times 2 + 1}}{2}\]
Multiplying 2 by 2, we get
\[ \Rightarrow F\left( 2 \right) = \dfrac{{4 + 1}}{2}\]
Adding the terms in the numerator, we get
\[ \Rightarrow F\left( 2 \right) = \dfrac{5}{2}\]
Therefore, we get
\[ \Rightarrow F\left( 2 \right) = 2.5\]
Substituting \[n = 2\] in the equation \[F\left( {n + 1} \right) = \dfrac{{2F\left( n \right) + 1}}{2}\], we get
\[ \Rightarrow F\left( {2 + 1} \right) = \dfrac{{2F\left( 2 \right) + 1}}{2}\]
\[ \Rightarrow F\left( 3 \right) = \dfrac{{2F\left( 2 \right) + 1}}{2}\]
Substituting \[F\left( 2 \right) = 2.5\] in the expression, we get
\[ \Rightarrow F\left( 3 \right) = \dfrac{{2 \times 2.5 + 1}}{2}\]
Multiplying 2 by \[2.5\], we get
\[ \Rightarrow F\left( 3 \right) = \dfrac{{5 + 1}}{2}\]
Adding the terms in the numerator, we get
\[ \Rightarrow F\left( 3 \right) = \dfrac{6}{2}\]
Therefore, we get
\[ \Rightarrow F\left( 3 \right) = 3\]
Substituting \[n = 3\] in the equation \[F\left( {n + 1} \right) = \dfrac{{2F\left( n \right) + 1}}{2}\], we get
\[ \Rightarrow F\left( {3 + 1} \right) = \dfrac{{2F\left( 3 \right) + 1}}{2}\]
\[ \Rightarrow F\left( 4 \right) = \dfrac{{2F\left( 3 \right) + 1}}{2}\]
Substituting \[F\left( 3 \right) = 3\] in the expression, we get
\[ \Rightarrow F\left( 4 \right) = \dfrac{{2 \times 3 + 1}}{2}\]
Multiplying 2 by 3, we get
\[ \Rightarrow F\left( 4 \right) = \dfrac{{6 + 1}}{2}\]
Adding the terms in the numerator, we get
\[ \Rightarrow F\left( 4 \right) = \dfrac{7}{2}\]
Therefore, we get
\[ \Rightarrow F\left( 4 \right) = 3.5\]
Therefore, we have
\[F\left( 1 \right) = 2\]
\[F\left( 2 \right) = 2.4\]
\[F\left( 3 \right) = 3\]
\[F\left( 4 \right) = 3.5\]
The sequence of \[F\left( 1 \right)\], \[F\left( 2 \right)\], \[F\left( 3 \right)\], \[F\left( 4 \right)\], … becomes 2, \[2.5\], 3, \[3.5\], …
We can observe that the sequence forms an arithmetic progression where 2 is the first term.
Here, the common difference \[ = 2.5 - 2 = 0.5\]
Also, we can observe that \[F\left( 1 \right)\] is the first term of the A.P., \[F\left( 2 \right)\] is the second term of the A.P., \[F\left( 3 \right)\] is the third term of the A.P., and \[F\left( 4 \right)\] is the fourth term of the A.P.
Therefore, we can conclude that
\[F\left( n \right)\] is the \[{n^{{\rm{th}}}}\] term of the A.P.
Now, we will find the value of \[F\left( {101} \right)\].
Since \[F\left( n \right)\] is the \[{n^{{\rm{th}}}}\] term of the A.P., \[F\left( {101} \right)\] is the \[{101^{{\rm{th}}}}\] term of the A.P.
The \[{n^{{\rm{th}}}}\] term of an A.P. is given by the formula \[{a_n} = a + \left( {n - 1} \right)d\], where \[a\] is the first term of the A.P. and \[d\] is the common difference.
We will use this formula to find the \[{101^{{\rm{th}}}}\] term of an A.P.
Substituting \[n = 101\], \[a = 2\], and \[d = 0.5\] in the formula for \[{n^{{\rm{th}}}}\] term of an A.P., we get
\[ \Rightarrow {a_{101}} = 2 + \left( {101 - 1} \right)\left( {0.5} \right)\]
Subtracting the terms in the parentheses, we get
\[ \Rightarrow {a_{101}} = 2 + 100\left( {0.5} \right)\]
Multiplying the terms of the expression, we get
\[ \Rightarrow {a_{101}} = 2 + 50\]
Adding 2 and 50, we get
\[ \Rightarrow {a_{101}} = 52\]
\[F\left( {101} \right)\] is the \[{101^{{\rm{th}}}}\] term of the A.P.
Therefore, we get
\[F\left( {101} \right) = 52\]
Therefore, we get the value of the expression \[F\left( {101} \right)\] as 52.
Thus, the correct option is option (b).
Note: We used the terms arithmetic progression and common difference in the solution.
An arithmetic progression is a series of numbers in which each successive number is the sum of the previous number and a fixed difference. The fixed difference is called the common difference.
We calculated the common difference by subtracting the first term from the second term. The common difference is the fixed difference between each successive term of an A.P.
Therefore, we get
Common difference \[ = \] Second term \[ - \] First term \[ = \] Third term \[ - \] Second term \[ = \] Fourth term \[ - \] Third term
We can observe that \[2.5 - 2 = 3 - 2.5 = 3.5 - 3\] all result in the same common difference \[0.5\].