The problem asks to prove the identity

$\sum_{k=1}^{n} \left( k^2 - k + 1 \right) = \frac{1}{3} n \left( n^2 + 2n \right)$

using the principle of Mathematical Induction.

Step 1: Base Case (n = 1)

For $n = 1$, the left-hand side (LHS) becomes:

$\sum_{k=1}^{1} \left( k^2 - k + 1 \right) = 1^2 - 1 + 1 = 1$

The right-hand side (RHS) for $n = 1$ is:

$\frac{1}{3} \times 1 \times (1^2 + 2 \times 1) = \frac{1}{3} \times 1 \times (1 + 2) = \frac{1}{3} \times 1 \times 3 = 1$

Since LHS = RHS for $n = 1$, the base case holds.

Step 2: Inductive Hypothesis

Assume that the statement is true for some $n = m$. That is, assume:

$\sum_{k=1}^{m} \left( k^2 - k + 1 \right) = \frac{1}{3} m \left( m^2 + 2m \right)$

Step 3: Inductive Step

We need to prove that the statement holds for $n = m + 1$, i.e.,

$\sum_{k=1}^{m+1} \left( k^2 - k + 1 \right) = \frac{1}{3} (m+1) \left( (m+1)^2 + 2(m+1) \right)$

Start with the left-hand side for $n = m + 1$:

$\sum_{k=1}^{m+1} \left( k^2 - k + 1 \right) = \sum_{k=1}^{m} \left( k^2 - k + 1 \right) + \left( (m+1)^2 - (m+1) + 1 \right)$

Using the inductive hypothesis:

$= \frac{1}{3} m \left( m^2 + 2m \right) + \left( (m+1)^2 - (m+1) + 1 \right)$

Simplify the expression for $(m+1)^2 - (m+1) + 1$:

$(m+1)^2 - (m+1) + 1 = m^2 + 2m + 1 - m - 1 + 1 = m^2 + m + 1$

Thus, the equation becomes:

$\frac{1}{3} m \left( m^2 + 2m \right) + (m^2 + m + 1)$

Now, combine these terms. First, express the second part as a fraction with denominator 3:

$= \frac{1}{3} m \left( m^2 + 2m \right) + \frac{3(m^2 + m + 1)}{3}$

Combine the terms over the common denominator:

$= \frac{1}{3} \left( m \left( m^2 + 2m \right) + 3(m^2 + m + 1) \right)$

Simplify the expression inside the parentheses:

$m \left( m^2 + 2m \right) = m^3 + 2m^2$ $3(m^2 + m + 1) = 3m^2 + 3m + 3$

Thus, we have:

$= \frac{1}{3} \left( m^3 + 2m^2 + 3m^2 + 3m + 3 \right)$ $= \frac{1}{3} \left( m^3 + 5m^2 + 3m + 3 \right)$

This is the expanded form of:

$\frac{1}{3} (m+1) \left( (m+1)^2 + 2(m+1) \right)$

Thus, the inductive step holds, and by the principle of mathematical induction, the identity is true for all $n \geq 1$.

Would you like more details or clarification on any step?

Here are 5 related questions for further practice:

1. Prove the formula for the sum of the first $n$ squares using induction.
2. Prove the identity $\sum_{k=1}^{n} k^3 = \left( \frac{n(n+1)}{2} \right)^2$ using mathematical induction.
3. Show that $2^n > n^2$ for all $n \geq 5$ using induction.
4. Verify the sum of an arithmetic progression formula using induction.
5. Use induction to prove that $n! > 2^n$ for $n \geq 4$.

Tip: When proving by induction, always carefully check both the base case and the transition from the $n = m$ to $n = m + 1$ case.