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

MATH220 April 2011

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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 $a_{n}=2^{n}-1\quad {\text{ for all }}n\in \mathbb {N} .$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Use induction!

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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.

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