Mathematical Induction (1 Viewer)

[rawker]

Can someone tell me if I'm so far going right - and how to finish it.

n
∑ 5r = 5/2 n (n+1) for all n≥1.
r=1

Let n=1

LHS= 5(1) = 5
RHS= 5/2 (1) ((1)+1) = 5

Assume the formula true for n=k.

5+10+15+...+5k=5/2 k (k+1)

Prove the formula true for n=k+1

..and thats where I'm lost.

 

[香港!]

"Assume the formula true for n=k.

5+10+15+...+5k=5/2 k (k+1)

Prove the formula true for n=k+1"
den put in the k+1 and try to prove it

5+10+15+...+5k+5(k+1)=5/2 (k+1) (k+1+1)

like that

 

[who_loves_maths]

Case for n = k +1:
i.e. to prove: 5 + 10 + ... + 5(k+1) = (5/2)(k+1)(k+2)

LHS = 5 + 10 + ... + 5k + 5(k +1)
= (5/2)k(k+1) + 5(k+1) .............................................. using your assumption.
= (5/2)(k+1)[k + 5*(2/5)] ........................................... factorising.
= (5/2)(k+1)(k+2)
= RHS

Hence, we've proven for the case n = k+1. So if the case for n = k is true, then n = k +1 must also be true. since it is true for n=1, then the general formula must be true for n = {2, 3, 4, ...} as well. i.e. true for all positive integral 'n'.
QED.

hope that helps

p.s. if the question says "or otherwise", just use the series&sequences ways, its much more expedient than induction.

 

[rawker]

LHS = 5 + 10 + ... + 5k + 5(k +1)
= (5/2)k(k+1) + 5(k+1) .............................................. using your assumption.
= (5/2)(k+1)[k + 5*(2/5)] ........................................... factorising.
= (5/2)(k+1)(k+2)
= RHS

I don't understand the second line... Where does 2/5 come from...
isnt it just (k+1)(k+5) i know your right-i just dont understand why....

 

[Slidey]

(5/2)k(k+1) + 5(k+1)
= 5(k+1)(k/2+1)
=(5/2)(k+1)(k+2)

edit: that's factorising - take out 5(k+1) and find a common denominator.

 

[100percent]

I don't understand the second line... Where does 2/5 come from...
isnt it just (k+1)(k+5) i know your right-i just dont understand why....

he factorised 5/2 out

 

[Riviet]

I'll break the proof down into smaller bits of working each line:
LHS=5+10+...+5k+5(k+1)
=(5/2)k(k+1)+5(k+1)
=5(k+1)(k/2+1) [taking a common factor of 5(k+1) out]
=5(k+1)(k/2+2/2) [1=2/2]
=5/2(k+1)(k+2) [taking a common factor of 1/2 out]
=RHS
Once you get better, you'll be able to skip a couple lines of working

 

[insert-username]

When you assume the formula true for n = k:

5+10+15+...+5k=5/2 k (k+1)

... you're saying that the equation is true for all numbers up to k. When you prove for n = k+1...

5+10+15+...+5k+5(k+1)=5/2 (k+1) (k+1+1)

... your assumption of 5+10+15+...+5k=5/2 k (k+1) equals the part in the bold above. Usually, the first part of proving true for n = k+1 for most induction questions is substituting in your assumption:

5+10+15+...+5k+5(k+1)=5/2*(k+1)(k+1+1)

But 5+10+15+...+5k = 5/2*k(k+1)

Therefore: 5/2*k(k+1) + 5(k+1) = 5/2*(k+1)(k+1+1)

From there, you can prove n = k+1. I hope that helps you.


I_F