Let the coefficients of x⁻¹ and x⁻³ in the expansion of (2x^(1/5) - 1/x^(1/5))¹⁵, x 0, be m and n

Q: Wait, why can't I just expand the whole thing? It's only power 15.

While power 15 is technically 'doable', it is a massive trap! In a JEE exam, time is your most precious currency. Expanding 16 terms is a recipe for calculation errors. The general term formula is your surgical tool—it lets you extract exactly what you need without the mess.

Q: I keep messing up the negative sign. How do I handle the $-x^{-1/5}$ part?

Treat the negative sign as a separate factor: $(-1)^k$. When you write your general term, always pull out the $(-1)^k$ immediately. It is a constant factor that depends on $k$. If $k$ is even, it vanishes; if $k$ is odd, it stays. Keep it separate, and you will never lose it.

Q: I got $k=10$ for $m$, so why isn't $r=10$?

That is a brilliant observation! You found the correct $k$ for $m$, but look at the structure of the final equation: ${}^{15}C_r \cdot 2^r$. If you plug in $r=10$, you get ${}^{15}C_{10} \cdot 2^{10}$. But our calculated value was ${}^{15}C_{10} \cdot 2^5$. The powers of 2 don't match! This is where the symmetry property ${}^{n}C_{k} = {}^{n}C_{n-k}$ saves the day, allowing us to transform ${}^{15}C_{10}$ into ${}^{15}C_{5}$ to match the power of 2.

Question

Let the coefficients of $x^{-1}$ and $x^{-3}$ in the expansion of $\left( 2x^{1/5} - \frac{1}{x^{1/5}} \right)^{15}$, $x > 0$, be $m$ and $n$ respectively. If $r$ is a positive integer such that $mn^2 = {}^{15}C_r \cdot 2^r$, then the value of $r$ is equal to____.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To solve the expansion of $\left( 2x^{\frac{1}{5}} - x^{-\frac{1}{5}} \right)^{15}$, we utilize the Binomial Theorem. The general term formula is given by:

$$T_{k+1} = {}^{n}C_{k} a^{n-k} b^{k}$$

Here, we define our variables as $a = 2x^{\frac{1}{5}}$, $b = -x^{-\frac{1}{5}}$, and $n = 15$. Substituting these into the formula, we obtain:

$$T_{k+1} = {}^{15}C_{k} \left( 2x^{\frac{1}{5}} \right)^{15-k} \left( -x^{-\frac{1}{5}} \right)^{k}$$

### Simplifying the General Term

To isolate the variables, we separate the constants from the powers of $x$. We extract the coefficients $2^{15-k}$ and $(-1)^k$ to the front:

$$T_{k+1} = {}^{15}C_{k} \cdot 2^{15-k} \cdot (-1)^k \cdot \left( x^{\frac{1}{5}} \right)^{15-k} \cdot \left( x^{-\frac{1}{5}} \right)^{k}$$

Applying the laws of exponents to the variable part, we combine the powers of $x$:

$$x^{\frac{15-k}{5} - \frac{k}{5}} = x^{\frac{15-2k}{5}}$$

### Finding the Coefficients

For the coefficient $m$ of $x^{-1}$, we set the exponent equal to $-1$:

$$\frac{15-2k}{5} = -1 \Rightarrow 15-2k = -5 \Rightarrow 2k = 20 \Rightarrow k = 10$$

Substituting $k=10$ into the constant part, we find:

$$m = {}^{15}C_{10} \cdot 2^{15-10} \cdot (-1)^{10} = {}^{15}C_{10} \cdot 2^5$$

For the coefficient $n$ of $x^{-3}$, we set the exponent equal to $-3$:

$$\frac{15-2k}{5} = -3 \Rightarrow 15-2k = -15 \Rightarrow 2k = 30 \Rightarrow k = 15$$

Substituting $k=15$ into the constant part, we find:

$$n = {}^{15}C_{15} \cdot 2^{15-15} \cdot (-1)^{15} = 1 \cdot 1 \cdot (-1) = -1$$

### Final Calculation

We are tasked with calculating $mn^2$. Since $n = -1$, it follows that $n^2 = 1$, and thus:

$$mn^2 = {}^{15}C_{10} \cdot 2^5$$

To match the form ${}^{15}C_r \cdot 2^r$, we apply the symmetry property of binomial coefficients, ${}^{n}C_{k} = {}^{n}C_{n-k}$:

$${}^{15}C_{10} = {}^{15}C_{15-10} = {}^{15}C_{5}$$

Substituting this back, we get ${}^{15}C_{5} \cdot 2^5$. Comparing this to ${}^{15}C_r \cdot 2^r$, we conclude that **$r = 5$**.

Enter Numerical Value:

Visualized Solution

The Binomial Expansion

General Term Formula

Substituting our Values

Simplifying the General Term

Isolating the Power of $x$

Finding Coefficient $m$

Calculating $m$

Finding Coefficient $n$

Calculating $n$

Evaluating $m n^{2}$

Comparing to Find $r$

The Sigma Insight: General Term and Middle Term

Analyzing the Setup

Simplifying the General Term

Finding the Coefficients

Final Calculation

Similar Questions

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Enter Numerical Value:

Visualized Solution

The Binomial Expansion

General Term Formula

Substituting our Values

Simplifying the General Term

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

Finding Coefficient .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

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

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

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

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } mn2

Comparing to Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

The Sigma Insight: General Term and Middle Term

Analyzing the Setup

Simplifying the General Term

Finding the Coefficients

Final Calculation

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Isolating the Power of $x$

Finding Coefficient $m$

Calculating $m$

Finding Coefficient $n$

Calculating $n$

Evaluating $m n^{2}$

Comparing to Find $r$

If the co-efficient of $x^{7}$ and $x^{8}$ in the expansion of $(2 + \frac{x}{3})^{n}$ are equal, then the value of $n$ is equal to ___

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