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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.

Prove the identity ∑_(k=1)^n (k^2 - k + 1) = (1/3)n(n^2 + 2n) for n = 1,2,3,... using the principle of Mathematical Induction.

Prove the summation identity ∑_(k=1)^n (k^2 - k + 1) = (1/3)n(n^2 + 2n) using the principle of Mathematical Induction. This solution includes detailed steps, covering algebraic manipulation and the induction hypothesis. Suitable for students in Grades 11-12.

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