Understanding the Binomial Theorem and General Term

The problem involves the binomial expansion of ${\left( {x + {a \over {{x^2}}}} \right)^n}$. A fundamental concept in solving such problems is the general term of a binomial expansion. For an expansion of the form $(X+Y)^n$, the general term, denoted as $T_{r+1}$, is given by the formula: $T_{r+1} = {}^nC_r X^{n-r} Y^r$ Here, ${}^nC_r$ is the binomial coefficient, calculated as $\frac{n!}{r!(n-r)!}$.

In our given expansion, $X=x$ and $Y=\frac{a}{x^2}$. Substituting these into the general term formula: $T_{r+1} = {}^nC_r (x)^{n-r} \left( \frac{a}{x^2} \right)^r$ To simplify this, we combine the terms involving $x$: $T_{r+1} = {}^nC_r x^{n-r} a^r (x^{-2})^r$ $T_{r+1} = {}^nC_r a^r x^{n-r-2r}$ $T_{r+1} = {}^nC_r a^r x^{n-3r}$ This simplified form of the general term is crucial as it directly provides the coefficient and the power of $x$ for any $(r+1)^{th}$ term. The coefficient of the $(r+1)^{th}$ term is ${}^nC_r a^r$.

Identifying Coefficients of Specific Terms

We are given information about the coefficients of the third, fourth, and fifth terms ($T_3, T_4, T_5$).

1. For the third term ($T_3$): Here, $r+1 = 3$, which means $r=2$. The coefficient of $T_3$ is ${}^nC_2 a^2$. The power of $x$ is $n-3(2) = n-6$. So, $T_3 = {}^nC_2 a^2 x^{n-6}$.
2. For the fourth term ($T_4$): Here, $r+1 = 4$, which means $r=3$. The coefficient of $T_4$ is ${}^nC_3 a^3$. The power of $x$ is $n-3(3) = n-9$. So, $T_4 = {}^nC_3 a^3 x^{n-9}$.
3. For the fifth term ($T_5$): Here, $r+1 = 5$, which means $r=4$. The coefficient of $T_5$ is ${}^nC_4 a^4$. The power of $x$ is $n-3(4) = n-12$. So, $T_5 = {}^nC_4 a^4 x^{n-12}$.

Setting Up and Solving Ratio Equations

The problem states that the coefficients of these terms are in the ratio $12:8:3$. This gives us two independent ratio equations to solve for $n$ and $a$.

Ratio of $T_3$ and $T_4$ Coefficients: We are given that the ratio of the coefficient of $T_3$ to the coefficient of $T_4$ is $\frac{12}{8} = \frac{3}{2}$. $\frac{\text{Coefficient of } T_3}{\text{Coefficient of } T_4} = \frac{{}^nC_2 a^2}{ {}^nC_3 a^3} = \frac{3}{2}$ We can simplify the binomial coefficient ratio using the identity $\frac{{}^nC_r}{ {}^nC_{r+1}} = \frac{r+1}{n-r}$. Here, $r=2$, so $\frac{{}^nC_2}{ {}^nC_3} = \frac{2+1}{n-2} = \frac{3}{n-2}$. Now substitute this back into our ratio equation: $\left( \frac{3}{n-2} \right) \cdot \left( \frac{a^2}{a^3} \right) = \frac{3}{2}$ $\frac{3}{a(n-2)} = \frac{3}{2}$ Dividing both sides by 3 and cross-multiplying gives us our first equation: $a(n-2) = 2 \quad \text{(Equation 1)}$

Ratio of $T_4$ and $T_5$ Coefficients: Next, we use the ratio of the coefficient of $T_4$ to the coefficient of $T_5$, which is $\frac{8}{3}$. $\frac{\text{Coefficient of } T_4}{\text{Coefficient of } T_5} = \frac{{}^nC_3 a^3}{ {}^nC_4 a^4} = \frac{8}{3}$ Again, using the identity $\frac{{}^nC_r}{ {}^nC_{r+1}} = \frac{r+1}{n-r}$. Here, $r=3$, so $\frac{{}^nC_3}{ {}^nC_4} = \frac{3+1}{n-3} = \frac{4}{n-3}$. Substitute this into the ratio equation: $\left( \frac{4}{n-3} \right) \cdot \left( \frac{a^3}{a^4} \right) = \frac{8}{3}$ $\frac{4}{a(n-3)} = \frac{8}{3}$ To solve for $a(n-3)$, we can cross-multiply: $4 \cdot 3 = 8 \cdot a(n-3)$ $12 = 8a(n-3)$ $a(n-3) = \frac{12}{8}$ $a(n-3) = \frac{3}{2} \quad \text{(Equation 2)}$

Solving for $n$ and $a$: We now have a system of two linear equations with two variables, $n$ and $a$:

1. $an - 2a = 2$
2. $an - 3a = \frac{3}{2}$

Subtract Equation 2 from Equation 1: $(an - 2a) - (an - 3a) = 2 - \frac{3}{2}$ $an - 2a - an + 3a = \frac{4}{2} - \frac{3}{2}$ $a = \frac{1}{2}$ Now substitute the value of $a = \frac{1}{2}$ back into Equation 1: $\frac{1}{2}(n-2) = 2$ Multiply both sides by 2: $n-2 = 4$ $n = 6$ So, we have found the values $n=6$ and $a=\frac{1}{2}$.

Finding the Term Independent of x

The term independent of $x$ is the term where the power of $x$ is zero. From our general term $T_{r+1} = {}^nC_r a^r x^{n-3r}$, we need to set the exponent of $x$ to zero: $n-3r = 0$ Substitute the value of $n=6$ we just found: $6-3r = 0$ $3r = 6$ $r = 2$ This means the third term ($T_{2+1} = T_3$) is the term independent of $x$.

Calculating the Final Value

Now, we substitute $n=6$, $a=\frac{1}{2}$, and $r=2$ into the coefficient part of the general term formula ${}^nC_r a^r$: $\text{Term independent of } x = {}^6C_2 \left( \frac{1}{2} \right)^2$ First, calculate ${}^6C_2$: ${}^6C_2 = \frac{6!}{2!(6-2)!} = \frac{6!}{2!4!} = \frac{6 \times 5}{2 \times 1} = 15$ Now, substitute this value: $\text{Term independent of } x = 15 \times \left( \frac{1}{4} \right)$ $\text{Term independent of } x = \frac{15}{4}$ The term independent of $x$ is $\frac{15}{4}$.

Tips and Common Mistakes

• Careful with General Term: Always write down and carefully simplify the general term $T_{r+1}$ first. Errors in exponents of $x$ are common.
• Binomial Coefficient Identities: Remember identities like $\frac{{}^nC_r}{ {}^nC_{r+1}} = \frac{r+1}{n-r}$ (or its reciprocal) to quickly simplify ratios of binomial coefficients. This saves time and reduces calculation errors.
• Solving System of Equations: When setting up ratio equations, ensure they are distinct and correctly represent the given information. Then, solve the system algebraically with precision.
• Term Independent of x: The condition for a term independent of $x$ is that the exponent of $x$ in the general term must be zero.
• Exact Values: In competitive exams like JEE, exact fractional answers are usually expected, not approximations (e.g., $3.75$ instead of $4$).

Summary

This problem demonstrates a classic application of the binomial theorem. By first determining the general term of the expansion, we systematically found the coefficients of the specified terms. Using the given ratios of these coefficients, we formed and solved a system of equations to find the values of $n$ and $a$. Finally, we used these values to identify the specific term independent of $x$ (where the power of $x$ is zero) and calculated its numerical value. The term independent of $x$ in the expansion is $\frac{15}{4}$.