Let [x] denote greatest integer less than or equal to x. If for n ∈ N, (1 - x + x³)^n = Σ (j=0 to 3n

Q: Wait, why do we use $x=1$ and $x=-1$? Can't we just expand the binomial using the multinomial theorem?

Expanding $(1 - x + x^3)^n$ using the multinomial theorem would be a massive, unnecessary calculation! The beauty of the substitution method is that it bypasses the need to find every single coefficient individually. We treat the polynomial as a function, and by choosing strategic inputs ($1$ and $-1$), we force the algebra to reveal the sums we need. It is the difference between brute force and elegant insight.

Q: The question asks for $\sum a_{2j} + 4 \sum a_{2j+1}$. Why is there a 4? Is it a trick?

It is not a trick, but it is a test of your confidence! Many students panic when they see a coefficient like '4' attached to the odd sum. Just treat it as a constant multiplier. Once you find the sum of odd coefficients, simply multiply that result by 4. Don't let the extra constant distract you from the core task of finding the sums.

Q: What if the polynomial was different? Does this method always work?

This is the 'Substitution Method' for coefficients, and it is a cornerstone of JEE algebra. It works for any polynomial $P(x)$. If you need the sum of all coefficients, set $x=1$. If you need the alternating sum, set $x=-1$. If you needed only even coefficients, you add the results and divide by 2. If you needed only odd coefficients, you subtract and divide by 2. This technique is universal!

Question

Let $[x]$ denote greatest integer less than or equal to $x$. If for $n \in \mathbb{N}$, $(1 - x + x^3)^n = \sum_{j=0}^{3n} a_j x^j$, then $\sum_{j=0}^{[\frac{3n}{2}]} a_{2j} + 4 \sum_{j=0}^{[\frac{3n-1}{2}]} a_{2j+1}$ is equal to :

Options:

(A) 2

(B) $2^{n-1}$

(C) 1

(D) $n$

Sigma Percentile · Accepted Answer

### The Art of Polynomial Deconstruction: A Journey into Coefficients

Welcome, future engineers. Today, we are going to dismantle a problem that, at first glance, looks like a nightmare of notation. You see a summation, you see a polynomial raised to the power of $n$, and you see a strange expression involving $a_{2j}$ and $a_{2j+1}$.

It is easy to feel overwhelmed. But I want you to take a deep breath. In the world of JEE Advanced, the most complex-looking problems are often hiding the most elegant, simple solutions.

We are not going to expand this polynomial. We are going to outsmart it.

### Phase 1: Decoding the Fear

Let us look at our given polynomial: 
$$P(x) = (1 - x + x^3)^n = \sum_{j=0}^{3n} a_j x^j$$

When you see $\sum_{j=0}^{[\frac{3n}{2}]} a_{2j}$, do not panic. What is $a_{2j}$? It is just the coefficient of the even powers of $x$.

If we wrote out the polynomial, it would look like $a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4 + \dots$. The sum $\sum a_{2j}$ is simply $a_0 + a_2 + a_4 + \dots$.

Similarly, $\sum a_{2j+1}$ is just $a_1 + a_3 + a_5 + \dots$. So, the question is not asking us to calculate individual coefficients. It is asking us to find the sum of even-indexed coefficients and add it to four times the sum of odd-indexed coefficients. We are looking for $S_{even} + 4S_{odd}$.

### Phase 2: The Substitution Magic

How do we isolate these sums without expanding the polynomial? This is where the 'Substitution Method' becomes our greatest weapon.

Think of $P(x)$ as a function. If we want the sum of all coefficients, we simply set $x = 1$. Why? Because $P(1) = a_0(1)^0 + a_1(1)^1 + a_2(1)^2 + \dots$, which simplifies to $a_0 + a_1 + a_2 + \dots$.

Now, what if we want the alternating sum? We set $x = -1$. Then $P(-1) = a_0 - a_1 + a_2 - a_3 + \dots$. This is the heartbeat of the solution.

### Phase 3: The Dance of Cancellation

First, let us find $P(1)$:
$$P(1) = (1 - 1 + 1^3)^n = (1)^n = 1$$

So, we have our first equation: $a_0 + a_1 + a_2 + a_3 + \dots = 1$. Let us call this Equation (1).

Next, let us find $P(-1)$:
$$P(-1) = (1 - (-1) + (-1)^3)^n = (1 + 1 - 1)^n = (1)^n = 1$$

So, we have our second equation: $a_0 - a_1 + a_2 - a_3 + \dots = 1$. Let us call this Equation (2).

Now, look at the beauty of these two equations. To find the sum of even coefficients ($S_{even} = a_0 + a_2 + \dots$), we add Equation (1) and Equation (2):
$$(a_0 + a_1 + a_2 + a_3 + \dots) + (a_0 - a_1 + a_2 - a_3 + \dots) = 1 + 1$$
$$2(a_0 + a_2 + a_4 + \dots) = 2$$
$$S_{even} = 1$$

To find the sum of odd coefficients ($S_{odd} = a_1 + a_3 + \dots$), we subtract Equation (2) from Equation (1):
$$(a_0 + a_1 + a_2 + a_3 + \dots) - (a_0 - a_1 + a_2 - a_3 + \dots) = 1 - 1$$
$$2(a_1 + a_3 + a_5 + \dots) = 0$$
$$S_{odd} = 0$$

### The Final Victory

We have arrived at the finish line. We needed to calculate $S_{even} + 4S_{odd}$. We found that $S_{even} = 1$ and $S_{odd} = 0$.

Substituting these values, we get:
$$1 + 4(0) = 1$$

And there it is. The complexity vanishes, leaving behind a simple, elegant integer. You didn't need to expand anything. You didn't need to struggle with binomial coefficients. You used the power of functional substitution to peel back the layers of the polynomial.

Keep this mindset—look for the structure, apply the theorem, and let the math do the work for you. The final answer is **1**.

Select Answer:

Visualized Solution

Analyzing the Given Expansion

Decoding the Required Expression

The Substitution Strategy

Substituting $x = 1$

Substituting $x = - 1$

Isolating Even Coefficients

Isolating Odd Coefficients

Final Calculation

The Sigma Insight: Properties of Binomial Coefficients

The Art of Polynomial Deconstruction

Phase 1

Phase 2

Phase 3

The Final Victory

Similar Questions

Let $n \in N$ and $[x]$ denote the greatest integer less than or equal to $x$ . If the sum of $(n + 1)$ terms $^{n} C_{0}, 3 \cdot^{n} C_{1}, 5 \cdot^{n} C_{2}, 7 \cdot^{n} C_{3}, \dots$ is equal to $2^{100} \cdot 101$ , then $2 [\frac{n - 1}{2}]$ is equal to ___

If $n \geq 2$ is a positive integer, then the sum of the series $^{n + 1} C_{2} + 2 (^{2} C_{2} +^{3} C_{2} +^{4} C_{2} + ... +^{n} C_{2})$ is :

Let $n$ be a positive integer and $(1 + x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + .... + a_{2 n} x^{2 n}$ . Show that $a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - .... + a_{2 n}^{2} = a_{n}$ .

Let $n$ be any positive integer. Prove that $\sum_{k = 0}^{m} \frac{( k 2 n - k )}{( n 2 n - k )} (2 n - 4 k + 1) 2^{n - 2 k} = \frac{( m n )}{( n - m 2 n - 2 m )} 2^{n - 2 m}$ for each non-negative integer $m \leq n$ .

$r$ and $n$ are positive integers $r > 1, n > 2$ and coefficient of $(r + 2)^{t h}$ term and $3 r^{t h}$ term in the expansion of $(1 + x)^{2 n}$ are equal, then $n$ equals

If the coefficients of $x^{4}, x^{5}$ and $x^{6}$ in the expansion of $(1 + x)^{n}$ are in the arithmetic progression, then the maximum value of $n$ is:

Find the sum of the series : $\sum_{r = 0}^{n} (- 1)^{r}^{n} C_{r} [\frac{1}{2 ^{r}} + \frac{3 ^{r}}{2 ^{2 r}} + \frac{7 ^{r}}{2 ^{3 r}} + \frac{1 5 ^{r}}{2 ^{4 r}} \dots up to m terms]$ .

If $\sum_{r = 0}^{2 n} a_{r} (x - 2)^{r} = \sum_{r = 0}^{2 n} b_{r} (x - 3)^{r}$ and $a_{k} = 1$ for all $k \geq n$ , then show that $b_{n} =^{2 n + 1} C_{n + 1}$ .

Let $n$ be a positive integer. Let $A = \sum_{k=0}^n (-1)^k {}^nC_k \left[ \left(\frac{1}{2}\right)^k + \left( rac{3}{4}\right)^k + \left(\frac{7}{8}\right)^k + \left(\frac{15}{16}\right)^k + \left(\frac{31}{32}\right)^k \right]$ . If $63 A = 1 - \frac{1}{2 ^{30}}$ , then $n$ is equal to ____

If $a_{n} = \sum_{r = 0}^{n} \frac{1}{^{n} C _{r}}$ , then $\sum_{r = 0}^{n} \frac{r}{^{n} C _{r}}$ equals