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

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The discussion focuses on proving the equation (2a+b) + (4a+b) + ... + (2na + b) = n(an + a + b) using mathematical induction for n ≥ 1. Participants clarify that the last term of the series is (2na + b), leading to specific sums for different values of n. For n=1, the sum simplifies to (2a+b), while for n=2 and n=3, it includes additional terms accordingly. There is an emphasis on ensuring that both sides of the equation are evaluated correctly when substituting n values. The conversation highlights the importance of understanding the series structure in the inductive proof process.
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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) ?
 
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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!
 
rooski said:

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.
 

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