Question

1. Prove that 12 + 22 + 32 + ⋯ + 𝑛𝑛2 = 𝑛𝑛(𝑛𝑛+1)6(2𝑛𝑛+1) , ∀𝑛𝑛 ∈ ℕ by using the principle of
mathematical induction.

Answer 1

To prove the given equation using mathematical induction, we need to follow two steps:
Step 1: Show that the equation holds true for the base case (n=1).
Step 2: Assume that the equation holds true for some arbitrary value k and then prove that it also holds true for (k+1).

Let's proceed with the proof:

Step 1: Base Case (n=1)
When n=1, the equation becomes:
12 = 1(1+1)/6(2*1+1)
12 = 1/6(2+1)
12 = 1/6(3)
12 = 1/2
12 = 1
The equation holds true for n=1.

Step 2: Inductive Step
Assume that the equation holds true for some arbitrary value k, i.e.,
12 + 22 + 32 + ⋯ + kk2 = kk(k+1)/6(2k+1)

Now, we need to prove that the equation also holds true for k+1.

We add (k+1)^3 to both sides of the equation:
12 + 22 + 32 + ⋯ + kk2 + (k+1)^3 = kk(k+1)/6(2k+1) + (k+1)^3

Simplifying the right side:
kk(k+1)/6(2k+1) + (k+1)^3
= (kk(k+1) + 6(2k+1)*(k+1)^3)/6(2k+1)
= (k^2(k+1) + 6(k+1)^4)/6(2k+1)
= (k+1)(k^2 + 6(k+1)^3)/6(2k+1)

Now we need to show that this expression is equal to (k+1)((k+1)+1)/6(2(k+1)+1), which is the equation we want to prove.

(k+1)((k+1)+1)/6(2(k+1)+1)
= (k+1)(k+2)/6(2k+3)

To prove that (k+1)(k^2 + 6(k+1)^3)/6(2k+1) = (k+1)(k+2)/6(2k+3), we can cancel out the common factors on the numerator and denominator.

(k+1)(k^2 + 6(k+1)^3)/6(2k+1)
= (k^2 + 6(k+1)^3)/(2k+1)
= [(k+1)(k^2 + 6(k+1)^3)]/[(k+1)(2k+1)]
= (k^2 + 6(k+1)^3)/(2k+1)

We can see that this expression is indeed equal to (k+1)(k+2)/6(2k+3), which means that the equation holds true for (k+1) as well.

By the principle of mathematical induction, we have proven that for all natural numbers n, 12 + 22 + 32 + ⋯ + nn2 = nn(n+1)/6(2n+1).