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A Sequence X 0 , X 1 , X 2 , X 3 , . . . is Defined by Letting X 0 = 5 a N D X K = 4 + X K − 1 for All Natural Number K . Show that X N = 5 + 4 N for All N ∈ N Using Mathematical Induction . - Mathematics

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Question

\[\text{ A sequence } x_0 , x_1 , x_2 , x_3 , . . . \text{ is defined by letting } x_0 = 5 and x_k = 4 + x_{k - 1}\text{  for all natural number k . } \]
\[\text{ Show that } x_n = 5 + 4n \text{ for all n }  \in N \text{ using mathematical induction .} \]

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Solution

\[\text{ Given: A sequence }  x_0 , x_1 , x_2 , x_3 , . . . \text{ is defined by letting }  x_0 = 5 \text{ and } x_k = 4 + x_{k - 1} \text{ for all natural number } k . \]

\[\text{ Let } P\left( n \right): x_n = 5 + 4n \text{ for all } n \in N . \]

\[\text{ Step I: For }  n = 0, \]

\[P\left( 0 \right): x_0 = 5 + 4 \times 0 = 5\]

\[\text{ So, it is true for }  n = 0 . \]

\[\text{ Step II: For } n = k, \]

\[\text{ Let } P\left( k \right): x_k = 5 + 4 \text{ k be true for some k } \in N . \]

\[\text{ Step III: For } n = k + 1, \]

\[P\left( k + 1 \right): x_{k + 1} = 4 + x_{k + 1 - 1} \]

\[ = 4 + x_k \]

\[ = 4 + 5 + 4k\]

\[ = 5 + 4\left( k + 1 \right)\]

\[\text{ So, it is also true for n }  = k + 1 . \]

\[\text{ Hence } , x_n = 5 + 4\text{ n for all  } n \in N .\]

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Chapter 12: Mathematical Induction - Exercise 12.2 [Page 29]

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RD Sharma Mathematics [English] Class 11
Chapter 12 Mathematical Induction
Exercise 12.2 | Q 48 | Page 29

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