Science:Math Exam Resources/Courses/MATH220/April 2011/Question 07 (a)

MATH220 April 2011

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[hide] Question 07 (a)
Let $\{a_{n}\}$ be a sequence of real numbers defined by $a_{1}=1\quad {\text{ and }}\quad a_{n+1}=2a_{n}+1\quad {\text{ for each }}n\in \mathbb {N} .$ Prove that $a_{n}=2^{n}-1\quad {\text{ for all }}n\in \mathbb {N} .$

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[show] Hint
Use induction!

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[show] Solution
We will prove this by induction. The first case, $n=1$ is easy to go through, indeed $\displaystyle 2^{1}-1=2-1=1=a_{1}$ so the first term of the sequence matches the general formula. You can do as many initial cases as you want, but remember, you have to at least one to get your argument started. Then comes the general case. We say: assume that the statement is true for all values of $n$ up to $m$ . We will show that the statement is also true for $n=m+1$ . By definition of the sequence ${a_{n}}$ , we know that $\displaystyle a_{m+1}=2a_{m}+1$ And we also assume that the statement is true up to $n=m$ hence $\displaystyle a_{m}=2^{m}-1$ Combining these two facts together, we obtain that $\displaystyle {\begin{aligned}a_{m+1}&=2a_{m}+1\\&=2(2^{m}-1)+1\\&=2^{m+1}-2+1\\&=2^{m+1}-1\end{aligned}}$ which is the claimed statement to be proved and so concludes our proof.