Key Concept: The General Term of a Binomial Expansion

The Binomial Theorem states that for any binomial $(a+b)^n$, the general term (or $(r+1)^{th}$ term), denoted as $T_{r+1}$, is given by: $T_{r+1} = {}^{n}{C_r} a^{n-r} b^r$ where ${}^{n}{C_r} = \frac{n!}{r!(n-r)!}$ is the binomial coefficient, and $r$ is an integer such that $0 \le r \le n$. A term is "independent of $x$" if the power of $x$ in that term is $0$ (i.e., $x^0$).

When finding the term independent of $x$ in a product of two expressions, say $P(x) \cdot Q(x)$, we consider each term in $P(x)$ and multiply it by a corresponding term in $Q(x)$ such that the sum of their powers of $x$ equals $0$.

Problem Setup

We need to find the term independent of $x$ in the expansion of $(1 - {x^2} + 3{x^3}){\left( {{5 \over 2}{x^3} - {1 \over {5{x^2}}}} \right)^{11}}$ This expression is a product of a polynomial $(1 - {x^2} + 3{x^3})$ and a binomial expansion ${\left( {{5 \over 2}{x^3} - {1 \over {5{x^2}}}} \right)^{11}}$.

Step 1: Determine the General Term of the Binomial Expansion

First, let's find the general term, $T_{r+1}$, of the binomial expansion ${\left( {{5 \over 2}{x^3} - {1 \over {5{x^2}}}} \right)^{11}}$. Here, we have $n=11$, $a = {5 \over 2}x^3$, and $b = - {1 \over {5{x^2}}}$. Applying the general term formula: $T_{r+1} = {}^{11}{C_r} \left( {{5 \over 2}{x^3}} \right)^{11-r} \left( {-{1 \over {5{x^2}}}} \right)^r$ To clearly separate the constant coefficients from the variable $x$ terms, we rewrite the expression: $T_{r+1} = {}^{11}{C_r} \left( {5 \over 2} \right)^{11-r} (x^3)^{11-r} \left( -{1 \over 5} \right)^r (x^{-2})^r$ Now, we combine the powers of $x$ using the rules $(x^a)^b = x^{ab}$ and $x^a \cdot x^b = x^{a+b}$: $T_{r+1} = {}^{11}{C_r} \left( {5 \over 2} \right)^{11-r} \left( -{1 \over 5} \right)^r x^{3(11-r) - 2r}$ Simplify the exponent of $x$: $3(11-r) - 2r = 33 - 3r - 2r = 33 - 5r$ So, the general term of the binomial expansion is: $T_{r+1} = {}^{11}{C_r} \left( {5 \over 2} \right)^{11-r} \left( -{1 \over 5} \right)^r x^{33-5r}$ This expression gives us the $(r+1)^{th}$ term of the binomial expansion, where $r$ must be an integer such that $0 \le r \le 11$.

Step 2: Identify Contributions to the Term Independent of x

The overall expression is $(1 - {x^2} + 3{x^3}) \cdot \left( \sum T_{r+1} \right)$. For a term to be independent of $x$, its total power of $x$ must be $0$. We consider each term from the polynomial $(1 - {x^2} + 3{x^3})$ multiplied by a term from the binomial expansion.

Case 1: When the term $1$ (from the polynomial) multiplies with $T_{r+1}$ For the product $1 \cdot T_{r+1}$ to be independent of $x$, the power of $x$ in $T_{r+1}$ must be $0$. So, we set the exponent of $x$ from $T_{r+1}$ to $0$: $33 - 5r = 0$ $5r = 33$ $r = \frac{33}{5}$ Since $r$ must be an integer, this case does not yield a term independent of $x$.

Case 2: When the term $-x^2$ (from the polynomial) multiplies with $T_{r+1}$ For the product $(-x^2) \cdot T_{r+1}$ to be independent of $x$ (i.e., $x^0$), the power of $x$ in $T_{r+1}$ must be $x^{-2}$ (because $x^{-2} \cdot x^2 = x^0$). So, we set the exponent of $x$ from $T_{r+1}$ to $-2$: $33 - 5r = -2$ $5r = 33 + 2$ $5r = 35$ $r = 7$ Since $r=7$ is an integer and falls within the valid range ($0 \le r \le 11$), this case contributes to the term independent of $x$. The coefficient for this term will be the product of the coefficient of $-x^2$ (which is $-1$) from the first polynomial and the coefficient of $x^{-2}$ (when $r=7$) from the binomial expansion. Contribution 1 = $(-1) \cdot \left[ {}^{11}{C_7} \left( {5 \over 2} \right)^{11-7} \left( -{1 \over 5} \right)^7 \right]$

Case 3: When the term $3x^3$ (from the polynomial) multiplies with $T_{r+1}$ For the product $(3x^3) \cdot T_{r+1}$ to be independent of $x$ (i.e., $x^0$), the power of $x$ in $T_{r+1}$ must be $x^{-3}$ (because $x^{-3} \cdot x^3 = x^0$). So, we set the exponent of $x$ from $T_{r+1}$ to $-3$: $33 - 5r = -3$ $5r = 33 + 3$ $5r = 36$ $r = \frac{36}{5}$ Since $r$ must be an integer, this case does not yield a term independent of $x$.

Step 3: Calculate the Contributing Term

Only Case 2, with $r=7$, contributes to the term independent of $x$. Let's calculate its value: Contribution 1 = $(-1) \cdot {}^{11}{C_7} \left( {5 \over 2} \right)^{4} \left( -{1 \over 5} \right)^7$

First, calculate the binomial coefficient ${}^{11}{C_7}$. We can use the property ${}^{n}{C_r} = {}^{n}{C_{n-r}}$: ${}^{11}{C_7} = {}^{11}{C_{11-7}} = {}^{11}{C_4}$ ${}^{11}{C_4} = \frac{11 \cdot 10 \cdot 9 \cdot 8}{4 \cdot 3 \cdot 2 \cdot 1} = \frac{11 \cdot (2 \cdot 5) \cdot (3 \cdot 3) \cdot (2^3)}{2^2 \cdot 3 \cdot 2 \cdot 1} = 11 \cdot 5 \cdot 3 \cdot 2 = 330$ Now, substitute this value and simplify the powers: $\left( {5 \over 2} \right)^4 = \frac{5^4}{2^4} = \frac{625}{16}$ $\left( -{1 \over 5} \right)^7 = \frac{(-1)^7}{5^7} = \frac{-1}{78125}$ Combine all parts to find the term: $\text{Term} = (-1) \cdot 330 \cdot \frac{625}{16} \cdot \frac{-1}{78125}$ The two negative signs multiply to a positive sign: $\text{Term} = 330 \cdot \frac{625}{16 \cdot 78125}$ We can simplify the powers of $5$: $625 = 5^4$ and $78125 = 5^7$. $\text{Term} = 330 \cdot \frac{5^4}{16 \cdot 5^7}$ Using the exponent rule $a^m / a^n = a^{m-n}$: $\text{Term} = 330 \cdot \frac{1}{16 \cdot 5^{7-4}}$ $\text{Term} = 330 \cdot \frac{1}{16 \cdot 5^3}$ Calculate $5^3 = 125$: $\text{Term} = 330 \cdot \frac{1}{16 \cdot 125}$ $\text{Term} = 330 \cdot \frac{1}{2000}$ $\text{Term} = \frac{330}{2000}$ Finally, simplify the fraction by dividing the numerator and denominator by $10$: $\text{Term} = \frac{33}{200}$

Common Mistakes to Avoid:

• Sign errors: Be meticulous with negative signs, especially when powers are involved. Remember that $(-1)^r$ alternates between $-1$ and $1$.
• Exponent rules: Carefully apply exponent rules such as $(x^a)^b = x^{ab}$ and $x^a \cdot x^b = x^{a+b}$. A common error is mixing these two.
• Integer values for r: Always ensure that the calculated value of $r$ is a non-negative integer within the range $[0, n]$ for the binomial coefficient to be valid.
• Simplifying fractions: Double-check calculations and simplify fractions accurately to their lowest terms.

Summary and Key Takeaway:

To find the term independent of $x$ in an expansion involving a product of a polynomial and a binomial, follow these steps:

1. Derive the general term ($T_{r+1}$) for the binomial expansion, expressing the power of $x$ as a function of $r$.
2. For each term in the polynomial, determine the required power of $x$ from the binomial's general term that would result in $x^0$ when multiplied.
3. Solve for $r$ for each case and discard non-integer or out-of-range values of $r$.
4. Calculate the coefficients for all valid $r$ values and sum them up to get the final term independent of $x$.

The term independent of $x$ in the given expansion is $\frac{33}{200}$.