Prove that 1
3 + 2
3 + 3
3 + ⋯ + 𝑛
3 = [
𝑛(𝑛+1)
2
]
2
using mathematical induction.
(ii) A team of 11 players is to be chosen from 15 members. In how many ways can this be 
done if (a) one particular player is always included? (b) two such players have always 
to be included?

Let's address both questions step by step.

### Part (i): Prove that 
\[
1^3 + 2^3 + 3^3 + \cdots + n^3 = \left( \frac{n(n+1)}{2} \right)^2
\]
using **mathematical induction**.

#### Step 1: Base Case (\( n = 1 \))
For \( n = 1 \), the left-hand side (LHS) is:
\[
1^3 = 1
\]
The right-hand side (RHS) is:
\[
\left( \frac{1(1+1)}{2} \right)^2 = \left( \frac{2}{2} \right)^2 = 1
\]
Since LHS = RHS, the base case holds.

#### Step 2: Inductive Hypothesis
Assume the formula holds for some arbitrary \( n = k \), i.e.,
\[
1^3 + 2^3 + 3^3 + \cdots + k^3 = \left( \frac{k(k+1)}{2} \right)^2
\]
This is the inductive hypothesis.

#### Step 3: Inductive Step
We must prove the formula holds for \( n = k+1 \), i.e., show that:
\[
1^3 + 2^3 + 3^3 + \cdots + (k+1)^3 = \left( \frac{(k+1)(k+2)}{2} \right)^2
\]

Start by adding \( (k+1)^3 \) to both sides of the inductive hypothesis:
\[
1^3 + 2^3 + \cdots + k^3 + (k+1)^3 = \left( \frac{k(k+1)}{2} \right)^2 + (k+1)^3
\]

Now, simplify the right-hand side:
\[
\left( \frac{k(k+1)}{2} \right)^2 + (k+1)^3 = (k+1)^2 \left( \frac{k^2}{4} + k+1 \right)
\]
\[
= (k+1)^2 \left( \frac{k^2 + 4k + 4}{4} \right) = (k+1)^2 \left( \frac{(k+2)^2}{4} \right)
\]
\[
= \frac{(k+1)^2(k+2)^2}{4} = \left( \frac{(k+1)(k+2)}{2} \right)^2
\]

Thus, the formula holds for \( n = k+1 \), completing the induction step.

### Conclusion:
By mathematical induction, we have proven that:
\[
1^3 + 2^3 + 3^3 + \cdots + n^3 = \left( \frac{n(n+1)}{2} \right)^2
\]
for all \( n \geq 1 \).

---

### Part (ii): A team of 11 players is to be chosen from 15 members.

#### (a) **If one particular player is always included**:

If one specific player is always included, we need to choose the remaining 10 players from the remaining 14 members. The number of ways to do this is given by the combination formula:
\[
\binom{14}{10} = \binom{14}{4} = \frac{14 \times 13 \times 12 \times 11}{4 \times 3 \times 2 \times 1} = 1001
\]
So, the number of ways to choose the team is \( 1001 \).

#### (b) **If two particular players are always included**:

If two specific players are always included, we need to choose the remaining 9 players from the remaining 13 members. The number of ways to do this is:
\[
\binom{13}{9} = \binom{13}{4} = \frac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1} = 715
\]
So, the number of ways to choose the team is \( 715 \).

---

### Do you want further details or explanations on any step? Here are 5 related questions for expanding the topic:

1. Can you prove the binomial formula used for combinations?
2. What are other examples of summation formulas you can prove using induction?
3. How does the combination formula change when more than two players are always included in the team?
4. How would you solve the team selection problem if there were additional restrictions (e.g., a maximum number of players from a subgroup)?
5. Can you prove the identity \( \sum_{i=1}^n i = \frac{n(n+1)}{2} \) using induction?

#### Tip: Always check both the base case and inductive step carefully when performing induction proofs.

Prove that 1^3 + 2^3 + 3^3 + ⋯ + n^3 = [n(n+1)/2]^2 using mathematical induction. (ii) A team of 11 players is to be chosen from 15 members. In how many ways can this be done if (a) one particular player is always included? (b) two such players have always to be included?

Explore how to prove the summation formula for cubes using mathematical induction: 1^3 + 2^3 + ⋯ + n^3 = [n(n+1)/2]^2. Also, learn to solve combinatorics problems related to team selection from a group of 15 members, with specific player inclusion conditions. Suitable for Grades 11-12.

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