1. Key Concepts and Formulas

This problem involves finding the constant term in the expansion of a trinomial raised to a power. The primary tools we will use are:

• Multinomial Theorem: For an expansion of the form $(a_1 + a_2 + \dots + a_k)^n$, the general term is given by: $\frac{n!}{n_1! n_2! \dots n_k!} a_1^{n_1} a_2^{n_2} \dots a_k^{n_k}$ where $n_1 + n_2 + \dots + n_k = n$ and $n_i \ge 0$ are integers.

• Binomial Theorem: A special case of the multinomial theorem for two terms $(a+b)^n$: $(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$ where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

Our goal is to find the term where the total power of $x$ is zero.

2. Problem Transformation: Simplifying the Expression

The given expression is ${\left( {2x + {1 \over {{x^7}}} + 3{x^2}} \right)^5}$. To make the process of identifying powers of $x$ simpler, we first rewrite the expression by taking out a common factor of $x^{-7}$ from each term inside the parenthesis. This strategy helps consolidate the $x$ terms.

$\left( 2x + x^{-7} + 3x^2 \right)^5$ Factor out $x^{-7}$ from each term within the parenthesis: $\left( x^{-7} \left( \frac{2x}{x^{-7}} + \frac{x^{-7}}{x^{-7}} + \frac{3x^2}{x^{-7}} \right) \right)^5$ $\left( x^{-7} \left( 2x^{1-(-7)} + 1 + 3x^{2-(-7)} \right) \right)^5$ $\left( x^{-7} \left( 2x^{8} + 1 + 3x^{9} \right) \right)^5$ Now, apply the power of $5$ to both parts: $(x^{-7})^5 \left( 1 + 2x^8 + 3x^9 \right)^5$ $x^{-35} \left( 1 + 2x^8 + 3x^9 \right)^5$ So, finding the constant term in the original expression is equivalent to finding the coefficient of $x^{35}$ in the expansion of $\left( 1 + 2x^8 + 3x^9 \right)^5$. We need to find terms from the expansion that, when multiplied by $x^{-35}$, result in $x^0$.

3. Applying the Binomial Theorem for Coefficient of $x^{35}$

Let's denote $A = (2x^8 + 3x^9)$. Then the expression becomes $(1 + A)^5$. We can expand this using the Binomial Theorem: $(1+A)^5 = \binom{5}{0} (1)^5 A^0 + \binom{5}{1} (1)^4 A^1 + \binom{5}{2} (1)^3 A^2 + \binom{5}{3} (1)^2 A^3 + \binom{5}{4} (1)^1 A^4 + \binom{5}{5} (1)^0 A^5$ We are looking for the term that will produce $x^{35}$. Let's analyze the powers of $x$ that each term $\binom{5}{k} A^k$ can generate:

• For $k=0$: $\binom{5}{0} A^0 = 1$ (constant term, no $x$).
• For $k=1$: $\binom{5}{1} A^1 = 5 (2x^8 + 3x^9)$. The highest power is $x^9$.
• For $k=2$: $\binom{5}{2} A^2 = 10 (2x^8 + 3x^9)^2 = 10(4x^{16} + 12x^{17} + 9x^{18})$. The highest power is $x^{18}$.
• For $k=3$: $\binom{5}{3} A^3 = 10 (2x^8 + 3x^9)^3$. The lowest power is $(x^8)^3 = x^{24}$. The highest power is $(x^9)^3 = x^{27}$.
• For $k=4$: $\binom{5}{4} A^4 = 5 (2x^8 + 3x^9)^4$. Let's examine this term in detail.
• For $k=5$: $\binom{5}{5} A^5 = 1 (2x^8 + 3x^9)^5$. The lowest power is $(x^8)^5 = x^{40}$.

From this analysis, only the term where $k=4$, i.e., $\binom{5}{4} A^4$, has the potential to generate $x^{35}$ because its powers range from $x^{32}$ to $x^{36}$.

4. Detailed Calculation for the Coefficient

We need to find the coefficient of $x^{35}$ in $\binom{5}{4} (2x^8 + 3x^9)^4$. First, calculate $\binom{5}{4} = \frac{5!}{4!1!} = 5$. Now, we need the coefficient of $x^{35}$ in $5(2x^8 + 3x^9)^4$. This means we need the coefficient of $x^{35}$ in $(2x^8 + 3x^9)^4$ and then multiply it by $5$.

Let's expand $(2x^8 + 3x^9)^4$ using the Binomial Theorem. Let $a = 2x^8$ and $b = 3x^9$. The general term is $\binom{4}{j} (2x^8)^{4-j} (3x^9)^j$. The power of $x$ in this term is $8(4-j) + 9j = 32 - 8j + 9j = 32 + j$. We want this power to be $35$: $32 + j = 35$ $j = 3$ Since $j=3$ is a valid value for $k$ (where $0 \le j \le 4$), this term exists.

Substitute $j=3$ back into the general term: $\binom{4}{3} (2x^8)^{4-3} (3x^9)^3$ $\binom{4}{3} (2x^8)^1 (3x^9)^3$ Calculate the coefficients: $\binom{4}{3} = \frac{4!}{3!1!} = 4$ The term becomes: $4 \times (2^1 x^8) \times (3^3 x^{27})$ $4 \times 2 \times x^8 \times 27 \times x^{27}$ $(4 \times 2 \times 27) x^{8+27}$ $(8 \times 27) x^{35}$ $216 x^{35}$ So, the coefficient of $x^{35}$ in $(2x^8 + 3x^9)^4$ is $216$.

Finally, we multiply this by the $\binom{5}{4}$ from the earlier expansion: Total coefficient of $x^{35} = 5 \times 216 = 1080$.

Thus, the constant term in the original expansion is $1080$.

Alternative approach from the provided solution: The step "${ }^{5} C_{4} \text { coefficient of } x^{3} \text { in }(2+3 x)^{4}$" is a clever simplification of the last part. Once we reach the stage of finding the coefficient of $x^{35}$ in $5(2x^8 + 3x^9)^4$: We can factor out $x^8$ from $(2x^8 + 3x^9)$: $5(x^8(2 + 3x))^4 = 5x^{32}(2+3x)^4$. To get $x^{35}$ from $5x^{32}(2+3x)^4$, we need to find the coefficient of $x^3$ in $(2+3x)^4$, and then multiply by $5$. Using the binomial theorem for $(2+3x)^4$, the general term is $\binom{4}{j} (2)^{4-j} (3x)^j = \binom{4}{j} 2^{4-j} 3^j x^j$. We need $x^j = x^3$, so $j=3$. The coefficient is $\binom{4}{3} 2^{4-3} 3^3 = \binom{4}{3} 2^1 3^3 = 4 \times 2 \times 27 = 216$. Then, multiply by $5$: $5 \times 216 = 1080$. This confirms our result and aligns with the provided solution's concise steps.

5. Tips and Common Mistakes

• Misinterpreting "constant term": A constant term means the term independent of $x$, i.e., the power of $x$ is 0.
• Incorrectly combining powers of x: When dealing with multiple terms inside the parenthesis, ensure all powers are correctly added/subtracted after multiplication.
• Forgetting binomial/multinomial coefficients: Don't just deal with the $x$ terms; remember the numerical coefficients from the theorem ($\binom{n}{k}$ or $n!/(p!q!r!)$).
• Algebraic errors in initial transformation: Be careful when factoring out terms or combining exponents like $x^{a}/x^{b} = x^{a-b}$.
• Not checking all possibilities: In complex expansions, multiple combinations of powers might lead to the desired exponent. A systematic approach (like varying $k$ or $q,r$) ensures no terms are missed.

6. Summary

To find the constant term in a complex trinomial expansion, we first simplify the expression by factoring out powers of $x$ to transform the problem into finding a specific power of $x$ in a simpler expansion. Then, we systematically apply the Binomial (or Multinomial) Theorem to identify the specific terms that contribute to the desired power of $x$. Careful calculation of coefficients and powers is crucial to arrive at the correct constant term. In this case, the constant term in the expansion of ${\left( {2x + {1 \over {{x^7}}} + 3{x^2}} \right)^5}$ is $\boxed{1080}$.