help with easy induction question (1 Viewer)

[xbimbo]

I'm quite bad at mathematical induction. What is the solution to this question?

1 + 2 + 3 + 4 +...+ n = n/2(n+1)

This is from the 3u Fitzpatrick textbook. Does anyone have worked solutions to exercise 23(c). I'm finding it hard

 

[LostAuzzie]

I'm quite bad at mathematical induction. What is the solution to this question?

1 + 2 + 3 + 4 +...+ n = n/2(n+1)

This is from the 3u Fitzpatrick textbook. Does anyone have worked solutions to exercise 23(c). I'm finding it hard

For n=1:
LHS = 1
RHS = 1/2 x 2
= 1
Therefore true for n = 1
Assume for n = k
ie: 1 + 2 + 3 + 4 + ... + k = k/2(k + 1)
Then for n = k + 1
LHS = 1 + 2 + 3 + 4 + ... + k + (k + 1)
= k/2(k + 1) + (k + 1)
RHS = (k+1)/2(k+2)
= (k^2)/2 + k/2 + k + 1
= k/2(k+1) + (k+1)
Therefore where true for n=k, true for n=k+1 etc etc etc

 

[word.]

test for n = 1
1 = 1/2(1 + 1) (true) therefore the statement is true for n = 1

assume true for n = k
i.e. 1+2+3+...+k=k/2(k+1)

proving statement for n = k+1
i.e. 1+2+3+...+k+(k+1) = (k+1)/2((k+1)+1)

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

LHS=(k² + k)/2 + k + 1
= (k² + k + 2k + 2)/2
= (k² + 3k + 2)/2
= (k+1)(k+2)/2 = RHS

therefore by induction etc etc.