Prove the following formulas: sumk=1ⁿ k2 = (n(n+1)(2n+1))/(6) and sumk=1ⁿ (2k−1)2 = (1)/(3)n(4n2-1)

LHS: $\sum_{k=1}^{1} k^2 = 1^2 = 1$.
RHS: $\frac{1(1+1)(2(1)+1)}{6} = \frac{1 \cdot 2 \cdot 3}{6} = 1$.
Since LHS = RHS, the formula holds for $n=1$.

Assume the formula holds for some positive integer $m$:

Prove the formula holds for $n=m+1$:

Substitute the inductive hypothesis:

Factor out $(m+1)$:

Factor the quadratic $2m^2+7m+6 = (m+2)(2m+3)$:

This is the RHS for $n=m+1$.
By mathematical induction, the formula is true for all $n \geq 1$.
The proof is complete.

: Prove that $\sum_{k=1}^{n} (2k-1)^2 = \frac{1}{3}n(4n^2-1)$.

Using $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$, $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$, and $\sum_{k=1}^{n} 1 = n$:

This matches the RHS.
The proof is complete.

: Prove that $n^3-n$ is divisible by 6 for all $n \geq 1$.

Using the difference of squares formula, $n^2-1 = (n-1)(n+1)$:

This expression represents the product of three consecutive integers.

• Divisibility by 2: In any set of three consecutive integers, at least one integer must be even. Therefore, their product $(n-1)n(n+1)$ is always divisible by 2.
• Divisibility by 3: In any set of three consecutive integers, exactly one integer must be a multiple of 3. Therefore, their product $(n-1)n(n+1)$ is always divisible by 3.

Since the product $(n-1)n(n+1)$ is divisible by both 2 and 3, and 2 and 3 are coprime, the product must be divisible by $2 \times 3 = 6$.
Thus, $n^3-n$ is divisible by 6 for all $n \geq 1$.
The proof is complete.