Let the coefficients of third, fourth and fifth terms in the expansion of (x + a/x²)^n, x ≠ 0, be in

Q: Wait, why do we ignore the $x$ terms when talking about coefficients?

Great question! In any term of the form $C \cdot x^k$, the coefficient is simply the scalar $C$. When the problem asks for the ratio of coefficients, it is asking for the ratio of the scalar parts. We isolate the $x$ part only to determine the power (to find the independent term), but for the ratio, we focus strictly on the constants.

Q: Is there a faster way than solving the system of equations?

The system is already quite fast! By dividing the equations, we reduce it to a simple linear equation in $n$. Trying to expand factorials like $\frac{n!}{(n-r)!r!}$ would lead to a messy, high-degree polynomial that is prone to calculation errors. Stick to the ratio property; it is the professional's shortcut.

Q: How do I know if I made a mistake in the ratio?

Check the symmetry. If $n$ comes out as a fraction or a negative number, you have likely flipped the ratio or misapplied the binomial property. Remember, $n$ must be a positive integer in a binomial expansion. If you get a non-integer, re-check your ratio setup.

Question

Let the coefficients of third, fourth and fifth terms in the expansion of $(x + \frac{a}{x^2})^n, x 
eq 0$, be in the ratio $12: 8: 3$. Then the term independent of $x$ in the expansion, is equal to ____.

Sigma Percentile · Accepted Answer

### The Binomial Symphony: Unlocking the Coefficients

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are decoding the DNA of a binomial expansion.

The problem asks us to dance with the coefficients of the third, fourth, and fifth terms of $(x + \frac{a}{x^2})^n$. It might look like a daunting wall of algebra, but it is a beautifully structured puzzle waiting for you to find the key.

### Phase 1: The Master Key

Every binomial expansion has a heartbeat: the general term. We define the $(r+1)$-th term as:

$$T_{r+1} = \binom{n}{r} x^{n-r} \left( \frac{a}{x^2} \right)^r$$

When we simplify this, we get:

$$T_{r+1} = \binom{n}{r} a^r x^{n-3r}$$

This is our master key. It tells us exactly what the coefficient is (the part with $\binom{n}{r}$ and $a^r$) and exactly what the power of $x$ is ($n-3r$). Keep this expression close; it is the foundation of everything we are about to do.

### Phase 2: The Ratio Trap

We are given the ratio of the coefficients of the third, fourth, and fifth terms as $12:8:3$. To find these, we plug in $r=2$ for the third term, $r=3$ for the fourth, and $r=4$ for the fifth.

This gives us coefficients of $\binom{n}{2} a^2$, $\binom{n}{3} a^3$, and $\binom{n}{4} a^4$.

Now, here is where many students stumble. Do not expand the factorials! Instead, use the elegant property of binomial coefficients:

$$\frac{\binom{n}{r}}{\binom{n}{r+1}} = \frac{r+1}{n-r}$$

This property is your best friend in JEE Advanced. It collapses the complexity into a simple linear form. For the ratio of the third and fourth terms, we set up:

$$\frac{\binom{n}{2} a^2}{\binom{n}{3} a^3} = \frac{12}{8} = \frac{3}{2}$$

Applying our property, this simplifies beautifully to:

$$\frac{3}{n-2} \cdot \frac{1}{a} = \frac{3}{2} \Rightarrow a(n-2) = 2$$

### Phase 3: The System of Equations

We repeat this logic for the ratio of the fourth and fifth terms:

$$\frac{\binom{n}{3} a^3}{\binom{n}{4} a^4} = \frac{8}{3}$$

Using our property again, this becomes:

$$\frac{4}{n-3} \cdot \frac{1}{a} = \frac{8}{3} \Rightarrow a(n-3) = \frac{3}{2}$$

Now, we have a system of two equations. The most elegant way to solve this is to divide the first by the second. The variable $a$ vanishes, leaving us with a simple linear equation in $n$.

Solving this, we find $n=6$. With $n$ in hand, finding $a$ is trivial: $a(6-2) = 2$, so $4a = 2$, which means **$a = \frac{1}{2}$**.

### Phase 4: The Independent Term

Finally, we reach the finish line. We need the term independent of $x$. This means the power of $x$ must be zero.

Looking back at our master key, the power of $x$ is $n-3r$. Setting $n-3r = 0$ with $n=6$, we get $6-3r=0$, so $r=2$.

This confirms that the third term is the one we are looking for. We calculate the value:

$$\binom{6}{2} a^2 = 15 \cdot \left( \frac{1}{2} \right)^2 = \frac{15}{4} = 3.75$$

Rounding to the nearest integer, we arrive at **$4$**.

Animated Solution for Mathematics - Binomial Theorem: Let the coefficients of third, fourth and fifth terms in the expansion of $(x + \frac{a}{x ^{2}})^{n}, x \neq = 0$ , be in the ratio $12 : 8 : 3$ . Then the term independent of $x$ in the expansion, is equal to ____.

Enter Numerical Value:

Visualized Solution

General Term Formula

Identifying Coefficients

Setting up Ratio $12 : 8$

Simplifying First Equation

Setting up Ratio $8 : 3$

Simplifying Second Equation

Solving for $n$

Solving for $a$

Term Independent of $x$

Final Result

The Sigma Insight: General Term and Middle Term

The Binomial Symphony

Phase 1

Phase 2

Phase 3

Phase 4

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Enter Numerical Value:

Visualized Solution

General Term Formula

Identifying Coefficients

Setting up Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 12:8

Simplifying First Equation

Setting up Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 8:3

Simplifying Second Equation

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Term Independent of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x