Induction Q (1 Viewer)

[bengosha60]

Prove by mathematical induction that:
(n+1) + (n+2) + ... + 2n = [n(3n+1)]/2 for n>=1

Taken from Fitzpatrick 3U 2ed 17.1 Q12

 

[photastic]

Does rearranging the assumption work for you, like i tried chucking the (k+1) over to the right but idk. ceeeebs

 

[braintic]

In a typical question, when n increases by 1, the first term is fixed, while one term is added at the end.
In this question, when n increase by 1, a term is LOST at the beginning, and TWO terms are added at the end.

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

LHS = [(n+1) + (n+2) + (n+3) + ... 2n] + (2n+1) + (2n+2) - (n+1) .... [adding (n+1) and subtracting it again]
= n(3n+1)/2 + (2n+1) + 2(n+1) - (n+1) (by assumption)
= n(3n+1)/2 + (2n+1) + (n+1)
= [n(3n+1) + 2(2n+1) + 2(n+1)] / 2
= (3n^2 + 7n + 4) / 2
= (n+1)(3n+4)/2
= RHS

 

[bengosha60]

How did you know that in this Q two terms are added at the end?

 

[Shadowdude]

Prove by mathematical induction that:
(n+1) + (n+2) + ... + 2n = [n(3n+1)]/2 for n>=1

Taken from Fitzpatrick 3U 2ed 17.1 Q12

Then for the n+1 case:

 

[braintic]

How did you know that in this Q two terms are added at the end?

Because substituting n+1 for n gives 2(n+1) which is 2n+2, which is two more than 2n

Since the pattern is that each term is increasing by 1, to get to 2n+2 we first need to add 2n+1.