FP1 Proof by induction of the standard result of r

When n=k

you sub in n for k

so you get

$\displaystyle\sum_{r=1}^k r = \frac{k}{2} (k+1)$

so when you use k+1 how would you do it?

so far i have

$\displaystyle\sum_{r=1}^(k+1) r = \frac{k+1}{2} (k+1+1)$??

add the result for k and the (k+1) term together

so in this case (k/2)(k+1)+(k+1)

add the result for k and the (k+1) term together

so in this case (k/2)(k+1)+(k+1)

so if i add the result for k then i have a 1 left right? but why add k+1 ???


ok so that's what you get when u do what you said....

THen add them together, and try and make it into what Sigma k+1 equals by factorising and stuff


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you add k+1 because if the summation formulae is to k+1 then its the sum to k plus the (k+1)th term, if that makes sense

When n=k

you sub in n for k

so you get

$\displaystyle\sum_{r=1}^k r = \frac{k}{2} (k+1)$

so when you use k+1 how would you do it?

so far i have

$\displaystyle\sum_{r=1}^{k+1} r = \frac{k+1}{2} (k+1+1)$??

Prove that it now works for n=k+1

$\displaystyle\sum_{r=1}^{k+1} r = \sum_{r=1}^{k} r + (k+1) = \frac{1}{2}k(k+1) + (k+1) = (k+1)[\frac{k}{2}+1] = \frac{1}{2}(k+1)(k+2)=\frac{k+1}{2}([k+1]+1)$

QED.

but shouldn't it be
$\displaystyle\sum_{r=1}^{k+1} r = \sum_{r=1}^k r + \sum_{r=1}^1 r$ ???

but shouldn't it be
$\displaystyle\sum_{r=1}^{k+1} r = \sum_{r=1}^k r + \sum_{r=1}^1 r$ ???

Of course not. That doesn't make sense. What you've shown says you are summing up integers from 1 up to k, and then adding an awkward 1 at the end.

You need to add the (k+1)th term of the sequence because you are showing it works for (1+2+3+4+5+...+k)+(k+1). Since you are summing up integers from 1, the (k+1)th term will simply be (k+1)

With proof by induction you always have to use your assumption. In this case we are assuming that $\displaystyle \sum_{r=1}^n r = \frac{1}{2} n(n+1)$. Using our assumption we have $\displaystyle \sum_{r=1}^{n+1} r = \frac{1}{2} n(n+1) + (n+1)$.
The way induction works is you assume that it works for some natural number $k$ and show that it implies that it works for the next natural number $k+1$. Then all you need to do is show that the result is indeed true for any value you pick and it completed the proof. Make sure you fully understand mathematical induction, I find that many people aren't convinced that it proves a result meaning that they do not properly understand it.

If you DO wish to express it as two different sums then:

$\displaystyle\sum_{r=1}^{k+1} r = \sum_{r=1}^k r + \sum_{r=k+1}^{k+1} r$

Notice the boundaries.

With proof by induction you always have to use your assumption. In this case we are assuming that $\displaystyle \sum_{r=1}^n r = \frac{1}{2} n(n+1)$. Using our assumption we have $\displaystyle \sum_{r=1}^{n+1} r = \frac{1}{2} n(n+1) + (n+1)$.
The way induction works is you assume that it works for some natural number $k$ and show that it implies that it works for the next natural number $k+1$. Then all you need to do is show that the result is indeed true for any value you pick and it completed the proof. Make sure you fully understand mathematical induction, I find that many people aren't convinced that it proves a result meaning that they do not properly understand it.

So is there anything i need to do to finalise the proof or does "therefore true for k+1 suffice"?

So is there anything i need to do to finalise the proof or does "therefore true for k+1 suffice"?

You usually need a final sentence saying:

The result is true for n=1; therefore true for n=2,3,4,... by induction. QED.

(QED is optional )