remolatg

2021-02-09

Consider the following recurrence relation:

Laaibah Pitt

Skilled2021-02-10Added 98 answers

Let the given recurrence relation as follows:

${a}_{n},=2\cdot {a}_{n-1}-3\text{}\text{with}\text{}{a}_{1}=5$

Let us consider the statement, for all natural number n, as follows

$P\left(n\right):{a}_{n},={2}^{n}+3,n\ge l$

We have to prove the statement P(n) by using Mathematical induction on n.

Show that the statement P(n) holds for$n=1.$

P(1) is easily seen to be true:

${a}_{1}=5={2}^{1}+3$

Inductive Step: Show that if P(k) holds then P$(k+1)$ holds.

Assume P(k) holds, that means,

${a}_{k}={2}^{k}+3$

Now we must have to show$P(k+1)$ holds, that is, to show

${a}_{k+1}={2}^{k+1}+3$

From the recurrece relation we have

${a}_{k+1}=2{a}_{k}-3$ [As given]

$=2({2}^{k}+3)-3$ [Induction Hypothesis]

$=2{a}^{k}+2\cdot 3-3$

$=2{a}_{k}+3$

Therefore, indeed,$P(k+1)$ is true.

Hence by mathematical induction, we conclude that P(n) is true for all

$n\ge 1.$

${a}_{n}={2}^{n}+3,n\ge 1$

Let us consider the statement, for all natural number n, as follows

We have to prove the statement P(n) by using Mathematical induction on n.

Show that the statement P(n) holds for

P(1) is easily seen to be true:

Inductive Step: Show that if P(k) holds then P

Assume P(k) holds, that means,

Now we must have to show

From the recurrece relation we have

Therefore, indeed,

Hence by mathematical induction, we conclude that P(n) is true for all

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