Let Cr denote the coefficient of xʳ in the binomial expansion of (1 + x)ⁿ, n ∈ N, 0 ≤ r ≤ n. If Pn =

Q: Why can't we just expand the binomials and solve it term by term?

Expanding term by term is a trap! It leads to a mess of factorials that are impossible to sum directly. The absorption identity is the elegant path designed to collapse the series into a single term. In JEE Advanced, we look for the 'shortcut' that avoids brute force.

Q: How do I know when to use the absorption identity?

Whenever you see a binomial coefficient $\binom{n}{r}$ multiplied by a rational function of $r$ (like $\frac{1}{r+1}$), that is your signal. It is a standard JEE pattern to absorb the denominator into the binomial coefficient to simplify the summation.

Q: Wait, why did the sum start at $s=1$ instead of $s=0$?

Great observation! When we substitute $s = r+1$, the lower limit of the sum changes. Since $r$ started at $0$, $s$ must start at $0+1=1$. Always check your limits after an index shift; it is the most common place to lose marks!

Q: Is there a trick to handling the $(-1)^{2n+1}$ term?

Yes! Always check the parity. Since $n$ is an integer, $2n+1$ is always odd. Therefore, $(-1)^{2n+1}$ is always $-1$. Recognizing this parity allows you to simplify complex expressions into simple constants instantly.

Question

Let $C_r$ denote the coefficient of $x^r$ in the binomial expansion of $(1 + x)^n, n \in \mathbb{N}, 0 \le r \le n$. If $P_n = C_0 - C_1 + \frac{2^2}{3}C_2 - \frac{2^3}{4}C_3 + \dots + \frac{(-2)^n}{n+1}C_n$, then the value of $\sum_{n=1}^{25} \frac{1}{P_{2n}}$ equals.

Options:

(A) 650

(B) 675

(C) 525

(D) 580

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The problem asks us to evaluate the series defined by:

$$P_n = C_0 - C_1 + \frac{2^2}{3}C_2 - \frac{2^3}{4}C_3 + \dots + \frac{(-2)^n}{n+1}C_n$$

When you first look at this expression, it is natural to feel intimidated. However, in the world of JEE Advanced, complexity is often just a mask for elegance.

### The Absorption Identity: Our Secret Weapon

The core of this problem lies in the term $\frac{1}{r+1} \binom{n}{r}$. Whenever you see a binomial coefficient divided by an index, your brain should immediately trigger the **Absorption Identity**.

This identity is defined as:

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

This is powerful because it takes the variable $r$ out of the denominator and absorbs it into the binomial coefficient itself. This is the key to unlocking the entire series.

### The Transformation

Let us substitute this identity into our summation. We rewrite $P_n$ as:

$$P_n = \sum_{r=0}^n (-2)^r \left[ \frac{1}{n+1} \binom{n+1}{r+1} ight]$$

Notice that $\frac{1}{n+1}$ is independent of $r$, so we can pull it out of the summation:

$$P_n = \frac{1}{n+1} \sum_{r=0}^n (-2)^r \binom{n+1}{r+1}$$

We are almost there, but there is a slight mismatch: the power of $-2$ is $r$, while the binomial coefficient has $r+1$. To fix this, we multiply and divide by $-2$ to get $(-2)^{r+1}$.

Now, let us perform an index shift. Let $s = r+1$. As $r$ goes from $0$ to $n$, $s$ goes from $1$ to $n+1$. Our expression becomes:

$$P_n = \frac{1}{-2(n+1)} \sum_{s=1}^{n+1} \binom{n+1}{s} (-2)^s$$

### The Final Victory

This looks exactly like the binomial expansion of $(1+x)^m$, where $x = -2$ and $m = n+1$. The full sum $\sum_{s=0}^{n+1} \binom{n+1}{s} (-2)^s$ would be $(1-2)^{n+1} = (-1)^{n+1}$.

Since our sum starts at $s=1$, we subtract the $s=0$ term, which is $\binom{n+1}{0}(-2)^0 = 1$. Thus, the sum is $(-1)^{n+1} - 1$.

Substituting this back, we get:

$$P_n = \frac{1 - (-1)^{n+1}}{2(n+1)}$$

When we evaluate $P_{2n}$, the term $(-1)^{2n+1}$ becomes $-1$, simplifying $P_{2n}$ to:

$$P_{2n} = \frac{1 - (-1)}{2(2n+1)} = \frac{2}{2(2n+1)} = \frac{1}{2n+1}$$

The final step is a simple summation: $\sum_{n=1}^{25} (2n+1)$. Using the sum of natural numbers formula, we calculate:

$$2 \left[ \frac{25 	imes 26}{2} ight] + 25 = 650 + 25 = \mathbf{675}$$

You have just navigated a complex series with pure logic. Keep this mindset, and no problem will ever be too difficult for you!

Select Answer:

Visualized Solution

Analyze the Series $P_{n}$

Identify the Binomial Identity

Substitute Identity into $P_{n}$

Factor Out Constants

Shift the Summation Index

Apply the Binomial Theorem

Simplify the Expression for $P_{n}$

Evaluate $P_{2 n}$

Set Up the Final Summation

Final Numerical Calculation

Summary and Key Takeaways

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Absorption Identity

The Transformation

The Final Victory

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Select Answer:

Visualized Solution

Analyze the Series .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Pn​

Identify the Binomial Identity

Substitute Identity into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Pn​