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This is a question from a test that I wrote and I'm wondering how do you solve it.

Prove by induction that

$$ \sum^n_{i=1}(i-1/2) = \frac{n^2}{2} $$

*Provide a Base Case, Inductive Hypothesis, and an Inductive Step

Julien
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user1766555
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1 Answers1

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Your base case will be the case $n=1$, which says that

$$\sum_{i=1}^1\left(i-\frac12\right)=\frac{1^2}2\tag{1}\;;$$

since both sides of $(1)$ simplify to $\frac12$, the base case is established. Your induction hypothesis will be that

$$\sum_{i=1}^m\left(i-\frac12\right)=\frac{m^2}2$$

for some particular $m\ge 1$, and from it you want to show that

$$\sum_{i=1}^{m+1}\left(i-\frac12\right)=\frac{(m+1)^2}2\;.\tag{2}$$

There’s a pretty standard trick for carrying out this kind of induction step: split the sum on the lefthand side of $(2)$ into the ‘old’ part — the sum from $1$ to $m$ — and the new term, and apply the induction hypothesis to the old part:

$$\begin{align*} \sum_{i=1}^{m+1}\left(i-\frac12\right)&=(m+1)-\frac12+\sum_{i=1}^m\left(i-\frac12\right)\\\\ &=m-\frac12+\frac{m^2}2&&\text{by the induction hypothesis}\\\\ &=\frac{m^2+2m+1}2\\\\ &=\frac{(m+1)^2}2\;, \end{align*}$$

which is exactly what we wanted. Since we’ve checked the base case and proved the induction step, we can now conclude that

$$\sum_{i=1}^n\left(i-\frac12\right)=\frac{n^2}2$$

for all integers $n\ge 1$.

Brian M. Scott
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  • A grader would probably also accept the standard induction proof for $\sum_{i=1}^{n}i = \frac{n(n+1)}{2}$, followed by the argument $\sum_{i=1}^{n}(i - \frac{1}{2}) = \sum_{i=1}^{n}i - \sum_{i=1}^{n}\frac{1}{2} = \frac{n(n+1)}{2} - \frac{n}{2} = \frac{n^2}{2}$ – colormegone Apr 18 '13 at 19:57
  • @RecklessReckoner: Probably is probably too strong: I’ve known a fair number of instructors who would not give full credit, arguing that it is not a proof by induction of the exact result. – Brian M. Scott Apr 18 '13 at 20:00
  • Fussy! Or if you're responding to the first time my comment appeared, I keep forgetting you can't hit 'Return' in a comment without exiting... – colormegone Apr 18 '13 at 20:02
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    @RecklessReckoner: (I was responding to the incomplete comment, but I knew what it was supposed to say.) I have mixed feelings. I wouldn’t ask for a specific method of proof in the first place, and I’d accept any legitimate argument. On the other hand, given that the instructor wanted specifically to see whether the student could carry you an induction that he hadn’t seen before, I can see not awarding full marks for a solution that technically doesn’t answer the question. In short, I’d not ask the question, but once it’s asked, I can see requiring that it actually be answered. – Brian M. Scott Apr 18 '13 at 20:09
  • Sorry to ask but could you show the step of: $ \sum_{i=1}^{m+1}\left(i-\frac12\right)=(m+1)-\frac12+\sum_{i=1}^m\left(i-\frac12\right) $ – user1766555 Apr 19 '13 at 04:25
  • @user1766555: I just separated the last term of the summation from the first $m$ terms. If $a_i=i-\frac12$, I converted $$\sum_{i=1}^{m+1}a_i=a_1+\ldots+a_m+a_{m+1}$$ to $$a_{m+1}+\sum_{i=1}^ma_i=a_{m+1}+(a_1+\ldots+a_m);.$$ – Brian M. Scott Apr 19 '13 at 05:56