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

MATH220 April 2011

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q1 (f)  •  Q1 (g)  •  Q1 (h)  •  Q1 (i)  •  Q1 (j)  •  Q2  •  Q3 (a)  •  Q3 (b)  •  Q4  •  Q5  •  Q6  •  Q7 (a)  •  Q7 (b)  •  Q7 (c)  •  Q8  •  Q9 (a)  •  Q9 (b)  •  Q9 (c)  •  Q10 (a)  •  Q10 (b)  •  

   Other MATH220 Exams

  •  April 2011  •  April 2005  •  December 2011  •  December 2010  •  December 2009  •  

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!

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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.

Click here for similar questions

MER QGH flag, MER QGQ flag, MER QGS flag, MER QGT flag, MER Tag Mathematical induction, Pages using DynamicPageList3 parser function, Pages using DynamicPageList3 parser tag