Let the coefficient of xʳ in the expansion of (x + 3)ⁿ⁻¹ + (x + 3)ⁿ⁻²(x + 2) + (x + 3)ⁿ⁻³(x + 2)² +

Q: Why is this a geometric progression?

Look at the terms. Each term is obtained by multiplying the previous one by $\frac{x+2}{x+3}$. That constant ratio is the definition of a geometric progression. Once you spot that, the problem becomes much easier.

Q: Why does $P(1)$ give the sum of coefficients?

If $P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_0$, then $P(1) = a_n(1)^n + a_{n-1}(1)^{n-1} + \dots + a_0 = a_n + a_{n-1} + \dots + a_0$, which is exactly the sum of all coefficients. It is a powerful shortcut in polynomial problems.

Q: Can I just expand everything using the binomial theorem?

You could, but it would be a nightmare of terms. The geometric progression approach is the elegant, JEE-style shortcut that avoids tedious expansion and leads directly to the answer.

Question

Let the coefficient of $x^r$ in the expansion of $(x + 3)^{n-1} + (x + 3)^{n-2}(x + 2) + (x + 3)^{n-3}(x + 2)^2 + \dots + (x + 2)^{n-1}$ be $\alpha_r$. If $\sum_{r=0}^n \alpha_r = \beta^n - \gamma^n, \beta, \gamma \in \mathbb{N}$, then the value of $\beta^2 + \gamma^2$ equals _______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The expression $(x+3)^{n-1} + (x+3)^{n-2}(x+2) + (x+3)^{n-3}(x+2)^2 + \dots + (x+2)^{n-1}$ appears daunting at first glance. However, by observing the pattern, we see that the power of $(x+3)$ decreases while the power of $(x+2)$ increases.

This is the classic signature of a **geometric progression**.

### The Engine of Simplification

We utilize the sum formula for a finite geometric progression, defined as:

$$S = a \frac{1-r^n}{1-r}$$

Here, the first term is $a = (x+3)^{n-1}$, the common ratio is $r = \frac{x+2}{x+3}$, and the total number of terms is $n$. Substituting these into the formula, we obtain:

$$S = (x+3)^{n-1} \frac{1 - \left( \frac{x+2}{x+3} \right)^n}{1 - \frac{x+2}{x+3}}$$

Focusing on the denominator, we simplify $1 - \frac{x+2}{x+3}$ as follows:

$$1 - \frac{x+2}{x+3} = \frac{(x+3) - (x+2)}{x+3} = \frac{1}{x+3}$$

The term $\frac{1}{x+3}$ in the denominator flips to the numerator, multiplying with $(x+3)^{n-1}$ to yield $(x+3)^n$.

### The Final Collapse

We now have the expression in the form:

$$S = (x+3)^n \left[ 1 - \frac{(x+2)^n}{(x+3)^n} \right]$$

Distributing $(x+3)^n$ inside the bracket, the terms cancel out with surgical precision. This leaves us with the simplified polynomial:

$$P(x) = (x+3)^n - (x+2)^n$$

### The Golden Rule of Coefficients

The problem asks for the sum of all coefficients. For any polynomial $P(x)$, the sum of its coefficients is found by substituting $x=1$.

This works because $1$ raised to any power remains $1$, effectively isolating the sum of the coefficients. We calculate:

$$P(1) = (1+3)^n - (1+2)^n = 4^n - 3^n$$

Comparing this to the given form $\beta^n - \gamma^n$, we identify $\beta = 4$ and $\gamma = 3$.

### Final Calculation

The question asks for the value of $\beta^2 + \gamma^2$. Substituting our identified values:

$$\beta^2 + \gamma^2 = 4^2 + 3^2 = 16 + 9 = 25$$

The final result is **25**.

Enter Numerical Value:

Visualized Solution

Analyze the Series Pattern

Identify G.P. Parameters

The G.P. Sum Formula

Substitution into Formula

Simplify the Denominator

Combine Outer Terms

Distribute and Simplify

Sum of Coefficients Property

Substitute $x = 1$

Compare with Given Form

Final Square Calculation

The Sigma Insight: Binomial Expansion for Positive Integral Index

Analyzing the Setup

The Engine of Simplification

The Final Collapse

The Golden Rule of Coefficients

Final Calculation

Similar Questions

The sum of the co-efficients of all even degree terms in $x$ in the expansion of $(x + x^{3} - 1)^{6} + (x - x^{3} - 1)^{6}, (x > 1)$ is equal to :

If $α$ and $β$ be the coefficient of $x^{4}$ and $x^{2}$ respectively in the expansion of $(x + x^{2} - 1)^{6} + (x - x^{2} - 1)^{6}$ , then :

In the expansion of $(1 + x) (1 - x^{2}) (1 + \frac{3}{x} + \frac{3}{x ^{2}} + \frac{1}{x ^{3}})^{5}, x \neq = 0$ , the sum of the coefficient of $x^{3}$ and $x^{- 13}$ is equal to _______.

If in the expansion of $(1 + x)^{p} (1 - x)^{q}$ the coefficients of $x$ and $x^{2}$ are 1 and -2, respectively, then $p^{2} + q^{2}$ is equal to:

If $α$ and $β$ be the coefficients of $x^{4}$ and $x^{2}$ respectively in the expansion of $(x + x^{2} - 1)^{6} + (x - x^{2} - 1)^{6}$ , then:

If in the expansion of $(1 + x)^{m} (1 - x)^{n}$ , the coefficients of $x$ and $x^{2}$ are 3 and $- 6$ respectively, then $m$ is

Let the sum of the coefficients of the first three terms in the expansion of $(x - \frac{3}{x ^{2}})^{n}, x \neq = 0, n \in N$ , be 376. Then the coefficient of $x^{4}$ is ______.

The sum of the co-efficients of all odd degree terms in the expansion of $(x + x^{3} - 1)^{5} + (x - x^{3} - 1)^{5}, (x > 1)$ is :

The coefficient of $x^{2012}$ in the expansion of $(1 - x)^{2008} (1 + x + x^{2})^{2007}$ is equal to