Proving Riemann sum rules via mathematical induction

Question

I have a question on my homework which states:

Prove the following rule via mathematical induction:

n ∑(3𝑖 − 2) i=1 = [n(3n − 1)]/2

I understand that induction refers to substituting in first something like n=1, then a random integer like k, then another like k+1, and if that is possible the rule is proved, but I do not understand how I can go around doing this? Is there anyone who could possibly explain?

Answer 1

You are correct on the process of induction. So let's roll with it.

Show true for n = 1.

n ∑(3𝑖 − 2) i=1 = [n(3n − 1)]/2

3(1) - 2 =? [1(3(1) - 1)]/2

3 - 2 =? [1(2)]/2

1 = 1√

The next step is to assume it is true for n = k. Let's find the pattern on the left.

n=1, (we've seen) ∑=1

n=2, ∑ = 1 + 3(2)-2 = 1 + 4

n=3, ∑ = 1 + 4 + 3(3)-2 = 1 + 4 + 7 (see the pattern?)

n=4, ∑ = 1 + 4 + 7 + 10

n=k, ∑ = 1 + 4 + 7 + 10 + ... + 3k - 2

right side

n=k, [k(3k − 1)]/2

Assume true for n = k

1 + 4 + 7 + 10 + ... + 3k - 2 = [k(3k − 1)]/2

The next step is to prove for n = k+1. The key thing is to remember to use the n=k!

Show true for n = k+1

1 + 4 + 7 + 10 + ... + 3k - 2 + 3(k+1) - 2 = [(k+1)(3(k+1)k − 1)]/2

let's replace the bold part on the left with [k(3k − 1)]/2 from the n=k.

[k(3k − 1)]/2 + 3(k+1) - 2 = [(k+1)(3(k+1)k − 1)]/2

See if you can take it from here.