Proof By Induction: Solving (2a+b) + ... + (2na + b) = n(an + a + b)

[rooski]

Homework Statement

give an inductive proof for all n >= 1

(2a+b) + (4a+b) + ... + (2na + b) = n(an + a + b)

The Attempt at a Solution

In order to start the method of inductoin i have to prove that f(1) holds true.

I am confused by the ... in this equation though.

If n = 1, then would the equation be (2a+b)+(4a+b)+(2a+b) = n(an + a + b) ?

 

[Mentallic]

No, the last term of the series is (2na+b) which means that for n=1, it is (2a+b), for n=2 it is (4a+b), n=3 is (6a+b) etc. You are shown n=1, n=2 and n=n

Oh and you need to substitute n=1 into the right hand side too!

 

[HallsofIvy]

Homework Statement

give an inductive proof for all n >= 1

(2a+b) + (4a+b) + ... + (2na + b) = n(an + a + b)

The Attempt at a Solution

In order to start the method of inductoin i have to prove that f(1) holds true.

I am confused by the ... in this equation though.

If n = 1, then would the equation be (2a+b)+(4a+b)+(2a+b) = n(an + a + b) ?

The last term in the sum is "2na+ b".

For n= 1, 2n= 2 so we have only the first term: 2a+ b.
For n= 2, 2n= 4 so we have the first two terms: (2a+ b)+ (4a+ b)
For n= 3, 2n= 6 so we have the first three terms (2a+ b)+ (4a+ b)+ (6a+ b)
etc.