Key Concepts and Formulas

This problem primarily relies on the Binomial Theorem for negative integer indices and the formula for the sum of a Geometric Progression.

1. Sum of a Geometric Progression: The sum of the first $n$ terms of a geometric progression with first term $a$ and common ratio $r$ is given by $S_n = \frac{a(1-r^n)}{1-r}$. A specific case relevant here is the sum $1 + x + x^2 + \dots + x^{n-1} = \frac{1-x^n}{1-x}$.

2. Binomial Theorem for Negative Integer Indices: For any real number $x$ with $|x| < 1$ and a positive integer $n$, the expansion of $(1-x)^{-n}$ is given by: $(1-x)^{-n} = 1 + nx + \frac{n(n+1)}{2!}x^2 + \frac{n(n+1)(n+2)}{3!}x^3 + \dots + \binom{n+r-1}{r}x^r + \dots$ The general term containing $x^r$ in the expansion of $(1-x)^{-n}$ is $\binom{n+r-1}{r}x^r$. This can also be written as $\binom{n+r-1}{n-1}x^r$.

Step-by-Step Solution

1. Simplify the base expression: The given expression is ${\left( {{{1 - {t^6}} \over {1 - t}}} \right)^3}$ Let's first analyze the term inside the parenthesis, $\frac{1-t^6}{1-t}$. This is a standard sum of a geometric progression where the first term is $1$, the common ratio is $t$, and there are $6$ terms (from $t^0$ to $t^5$). Therefore, we can simplify: $\frac{1-t^6}{1-t} = 1 + t + t^2 + t^3 + t^4 + t^5$ Explanation: We simplify this expression to make the subsequent binomial expansion more manageable. Recognizing this as a geometric series sum is a crucial first step.

Now, the original expression becomes: $(1 + t + t^2 + t^3 + t^4 + t^5)^3$ Explanation: By simplifying the base, we transform the problem into finding the coefficient of $t^4$ in the expansion of a polynomial raised to the power of 3.

Alternative Approach (using the given solution's initial step): If we strictly follow the original solution's factorization, we would rewrite the expression as: $(1 - t^6)^3 (1 - t)^{-3}$ Explanation: This step separates the numerator and the denominator, allowing us to apply the Binomial Theorem to each part.

2. Expand each factor (if using the alternative approach):

First, let's expand $(1 - t^6)^3$ using the binomial expansion formula $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$: $(1 - t^6)^3 = 1^3 - 3(1^2)(t^6) + 3(1)(t^6)^2 - (t^6)^3$ $(1 - t^6)^3 = 1 - 3t^6 + 3t^{12} - t^{18}$ Explanation: We expand this polynomial to identify all terms that could potentially contribute to the $t^4$ term when multiplied by the expansion of $(1-t)^{-3}$.

Next, let's expand $(1 - t)^{-3}$ using the Binomial Theorem for negative integer indices. Here, $x=t$ and $n=3$. The general term is $\binom{n+r-1}{r}t^r = \binom{3+r-1}{r}t^r = \binom{r+2}{r}t^r$. So, $(1 - t)^{-3} = \binom{2}{0} + \binom{3}{1}t + \binom{4}{2}t^2 + \binom{5}{3}t^3 + \binom{6}{4}t^4 + \dots$ Explanation: We need the general form of the expansion of $(1-t)^{-3}$ to multiply it with the terms from the first expansion.

3. Identify terms contributing to $t^4$:

Now we need to find the coefficient of $t^4$ in the product of $(1 - 3t^6 + 3t^{12} - t^{18})$ and $(1 - t)^{-3}$. We look for combinations of terms from each expansion whose powers of $t$ sum to $4$.

Consider the terms from $(1 - 3t^6 + 3t^{12} - t^{18})$:

• The term $1$: To get $t^4$, we need to multiply $1$ by the $t^4$ term from $(1 - t)^{-3}$.
• The term $-3t^6$: If we multiply this by any term from $(1 - t)^{-3}$, the resulting power of $t$ will be $t^6 \cdot t^k = t^{6+k}$, which will always be greater than $4$ (since $k \ge 0$). Thus, this term and any subsequent terms ($3t^{12}$, $-t^{18}$) will not contribute to the coefficient of $t^4$.

Therefore, the coefficient of $t^4$ comes solely from the product of the constant term $1$ from $(1 - t^6)^3$ and the coefficient of $t^4$ from $(1 - t)^{-3}$. Explanation: This step highlights the efficiency of recognizing which terms are relevant. Higher-order terms in the first factor can be ignored as they cannot produce $t^4$ when multiplied by non-negative powers of $t$.

4. Calculate the required coefficient:

We need the coefficient of $t^4$ in the expansion of $(1 - t)^{-3}$. Using the general term $\binom{r+2}{r}t^r$, for $r=4$, the coefficient is $\binom{4+2}{4} = \binom{6}{4}$.

Now, calculate the binomial coefficient $\binom{6}{4}$: $\binom{6}{4} = \frac{6!}{4!(6-4)!} = \frac{6!}{4!2!} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{(4 \times 3 \times 2 \times 1)(2 \times 1)}$ $\binom{6}{4} = \frac{6 \times 5}{2 \times 1} = \frac{30}{2} = 15$ Explanation: This is the direct application of the binomial coefficient formula. Remember that $\binom{n}{k} = \binom{n}{n-k}$, so $\binom{6}{4} = \binom{6}{2}$, which is often quicker to calculate.

So, the coefficient of $t^4$ in the entire expression is $1 \times 15 = 15$.

Tips and Common Mistakes

• Don't forget the base simplification: Always check if the base of the binomial expression can be simplified (e.g., using geometric series formulas) before expanding. This can often lead to a much simpler expansion.
• Careful with signs: When using the binomial theorem for $(1-x)^{-n}$, ensure correct signs. The formula $\binom{n+r-1}{r}x^r$ correctly handles $(1-x)^{-n}$ for positive $x$.
• Recognize irrelevant terms: In a product of expansions, terms with powers higher than the target power can often be ignored, simplifying calculations.
• Binomial coefficient properties: Remember properties like $\binom{n}{k} = \binom{n}{n-k}$ to simplify calculations. For example, $\binom{6}{4}$ is easier to compute as $\binom{6}{2}$.

Summary and Key Takeaway

This problem effectively tests the understanding and application of the Binomial Theorem for negative integer indices, alongside the recognition of geometric series sums. The key to solving such problems efficiently lies in strategically simplifying the expression first and then carefully identifying only those terms from the expansion that contribute to the desired power. The coefficient of $t^4$ in the given expansion is $\boxed{\text{15}}$.