# Consider the following recurrence relation: a_n=2 cdot a_n-1-3 with a_1=5

Consider the following recurrence relation:

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Laaibah Pitt
Let the given recurrence relation as follows:

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 $\left(k+1\right)$ holds.
Assume P(k) holds, that means,
${a}_{k}={2}^{k}+3$
Now we must have to show $P\left(k+1\right)$ 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]
$=2\left({2}^{k}+3\right)-3$ [Induction Hypothesis]
$=2{a}^{k}+2\cdot 3-3$
$=2{a}_{k}+3$
Therefore, indeed, $P\left(k+1\right)$ is true.
Hence by mathematical induction, we conclude that P(n) is true for all
$n\ge 1.$
${a}_{n}={2}^{n}+3,n\ge 1$