A sequence (x_0, x_1, x_2, x_3, ......) is defined by letting $x_{0} = 5$ and $x_{k} = 4 + x_{k - 1}$ for all natural numbers k. Show that xn = 5 + 4n for all n $ϵ$ N using mathematical induction.

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Solution

Consider the given statement

P(n) : $b_{n} = 5 + 4 n$ , for all natural numbers

given that $b_{0} = 5$ and $b_{k} = 4 + b_{k - 1}$

Step I P(1) is true.

P(1) : $b_{1} = 5 + 4 \times 1 = 9$

As $b_{0} = 5, b_{1} = 4 + b_{0} = 4 + 5 = 9$

Hence, P(1) is true.

Step II Now, assume that P(n) is true for n = k

P(k) : $b_{k} = 5 + 4 k$

Step III Now, to prove P(k + 1) is true, we have to show that

$∴$ $P (k + 1) : b_{k + 1} = 5 + 4 (k + 1)$

$b_{k + 1} = 4 + b_{k + 1 - 1}$

$= 4 + b_{k}$

$= 4 + 5 + 4 k = 5 + 4 (k + 1)$

So, by the mathematical induction P(k + 1) is ture whanever P(k) is true, hence P(n) is true.

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A sequence (x_0, x_1, x_2, x_3, ......) is defined by letting x0=5 and xk=4+xk−1 for all natural numbers k. Show that xn = 5 + 4n for all n ϵ N using mathematical induction.

A sequence (x_0, x_1, x_2, x_3, ......) is defined by letting $x_{0} = 5$ and $x_{k} = 4 + x_{k - 1}$ for all natural numbers k. Show that xn = 5 + 4n for all n $ϵ$ N using mathematical induction.