Prove that C₀ - 2² C₁ + 3² C₂ - .... + (-1)ⁿ (n + 1)² Cn = 0, n 2, where Cr = nCr.

Q: Wait, why can't I just differentiate $(1+x)^n$ twice to get the $r^2$ term? Is that not faster?

That is a brilliant observation! Yes, differentiating $(1+x)^n$ twice is a perfectly valid and powerful method. However, in the heat of a JEE Advanced exam, differentiation can sometimes lead to messy algebra or sign errors when dealing with the alternating $(-1)^r$ term. The 'falling factorial' method I showed you is often more robust because it keeps the summation structure intact and avoids the potential pitfalls of product rules and chain rules. Both paths lead to the same truth, but choose the one where you feel most confident.

Q: Why do we need the condition $n > 2$?

Great catch! When we use the identity $r(r-1) \cdot {}^nC_r = n(n-1) \cdot {}^{n-2}C_{r-2}$, the binomial coefficient ${}^{n-2}C_{r-2}$ only exists if $n-2 \ge 0$, which means $n \ge 2$. If $n$ were smaller, the term would be undefined or zero. We specify $n > 2$ to ensure that all our binomial coefficients are well-defined and that the series has enough terms to actually 'collapse' to zero. It is all about maintaining mathematical rigor.

Q: Is there a shortcut to memorizing these identities?

Don't memorize—internalize! In JEE, the shortcut is recognizing the pattern. Whenever you see a polynomial in $r$ multiplied by ${}^nC_r$, immediately think: 'I need to reduce this to linear and constant terms to match the binomial coefficients.' Once you see $(r+1)^2$, your brain should automatically trigger the expansion $r(r-1) + 3r + 1$. Practice makes this intuition second nature, not rote memorization.

Question

Prove that $C_0 - 2^2 C_1 + 3^2 C_2 - .... + (-1)^n (n + 1)^2 C_n = 0, n > 2$, where $C_r = ^nC_r$.

Sigma Percentile · Accepted Answer

### The Symphony of Binomial Coefficients

Welcome, my dear student. Today, we are not just solving a problem; we are conducting an orchestra. We are looking at the expression $S = \sum_{r=0}^n (-1)^r (r+1)^2 {}^nC_r$.

At first glance, it looks like a chaotic mess of alternating signs and quadratic growth. But in the world of JEE Advanced, chaos is often just order waiting to be discovered.

### Phase 1: The Algebraic Surgery

We start by staring down the enemy: $(r+1)^2$. This term is the obstacle preventing us from using the simple binomial expansion of $(1-1)^n$.

If we had just ${}^nC_r$, the sum would be zero. If we had $r \cdot {}^nC_r$, we could use the absorption identity. But $(r+1)^2$ is a quadratic, so we must perform surgery.

We expand $(r+1)^2$ into $r^2 + 2r + 1$. To make this play nicely with the binomial coefficient ${}^nC_r$, we use the 'falling factorial' trick: $r^2 = r(r-1) + r$.

Why? Because $r(r-1)$ is the key to unlocking the second-order absorption identity. Our expression becomes $r(r-1) + r + 2r + 1$, which simplifies beautifully to $r(r-1) + 3r + 1$.

Now, we have a linear combination of terms that we can handle individually.

### Phase 2: The Power of Linearity

Now, we invoke the linearity of summation. We split our massive, intimidating sum into three manageable parts:

$$S = \sum_{r=0}^n (-1)^r r(r-1) {}^nC_r + 3 \sum_{r=0}^n (-1)^r r {}^nC_r + \sum_{r=0}^n (-1)^r {}^nC_r$$

Imagine these as three separate soldiers. The first one is the most complex, but it is also the most elegant. The second one is a standard application of the absorption identity, and the third one is the classic binomial expansion.

### Phase 3: The Absorption Identity

This is where the magic happens. Recall the fundamental identity: $r \cdot {}^nC_r = n \cdot {}^{n-1}C_{r-1}$. If we apply this twice, we get:

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

Look at the first sum: $\sum_{r=0}^n (-1)^r r(r-1) {}^nC_r$. Substituting our identity, it becomes:

$$n(n-1) \sum_{r=2}^n (-1)^r {}^{n-2}C_{r-2}$$

If we let $k = r-2$, this transforms into $n(n-1) \sum_{k=0}^{n-2} (-1)^{k+2} {}^{n-2}C_k$. Since $(-1)^{k+2} = (-1)^k$, this is just $n(n-1) (1-1)^{n-2}$, which is $0$.

The first soldier has fallen, and he has taken the complexity with him!

### Phase 4: The Grand Finale

We repeat this logic for the second sum: $3 \sum_{r=0}^n (-1)^r r {}^nC_r$. Using $r \cdot {}^nC_r = n \cdot {}^{n-1}C_{r-1}$, this becomes:

$$3n \sum_{r=1}^n (-1)^r {}^{n-1}C_{r-1}$$

Again, shifting the index, we get $3n(1-1)^{n-1}$, which is also $0$. Finally, the third sum is simply $\sum_{r=0}^n (-1)^r {}^nC_r$, which is the definition of $(1-1)^n$, which is $0$.

When we combine them, $0 + 0 + 0 = 0$. We have proven the identity.

The beauty here is not just in the result, but in the realization that no matter how complex the polynomial in $r$ is, as long as it is a polynomial, we can always decompose it into falling factorials and reduce it to zero. You have mastered a core technique of combinatorics; keep this tool in your arsenal, and you will be ready for whatever the exam throws at you.

The final result is **$0$** (for $n \ge 2$).

Animated Solution for Mathematics - Binomial Theorem: Prove that $C_{0} - 2^{2} C_{1} + 3^{2} C_{2} - .... + (- 1)^{n} (n + 1)^{2} C_{n} = 0, n > 2$ , where $C_{r} =^{n} C_{r}$ .

Visualized Solution

Identifying the General Term $T_{r}$

Expanding the Quadratic Term $(r + 1)^{2}$

Using Falling Factorials for $r^{2}$

Splitting the Summation

Property of Binomial Coefficients

Evaluating the First Sum

Evaluating the Second Sum

Evaluating the Third Sum

Combining the Results

The Sigma Insight: Properties of Binomial Coefficients

The Symphony of Binomial Coefficients

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

Given that $C_{1} + 2 C_{2} x + 3 C_{3} x^{2} + .... + 2 n C_{2 n} x^{2 n - 1} = 2 n (1 + x)^{2 n - 1}$ where $C_{r} = \frac{( 2 n )!}{r ! ( 2 n - r )!}, r = 0, 1, 2, ...., 2 n$ . Prove that $C_{1}^{2} - 2 C_{2}^{2} + 3 C_{3}^{2} - .... - 2 n C_{2 n}^{2} = (- 1)^{n} n^{n} C_{n}$ .

Prove that $\sum_{r=1}^k (-3)^{r-1} ^{3n}C_{2r-1} = 0$ , where $k = (3 n) /2$ and $n$ is an even positive integer.

Using mathematical induction, prove that $^{m} C_{0}^{n} C_{k} +^{m} C_{1}^{n} C_{k - 1} + .... +^{m} C_{k}^{n} C_{0} =^{(m + n)} C_{k}$ , where $m, n, k$ are positive integers, and $^{p} C_{q} = 0$ for $p < q$ .

Prove that $2^{k} (0 n) (k n) - 2^{k - 1} (1 n) (k - 1 n - 1) + 2^{k - 2} (2 n) (k - 2 n - 2) - .... + (- 1)^{k} (k n) (0 n - k) = (k n)$

Given $s_{n} = 1 + q + q^{2} + .... + q^{n}$ ; $S_{n} = 1 + \frac{q + 1}{2} + (\frac{q + 1}{2})^{2} + .... + (\frac{q + 1}{2})^{n}, q \neq = 1$ Prove that $^{n + 1} C_{1} +^{n + 1} C_{2} s_{1} +^{n + 1} C_{3} s_{2} + .... +^{n + 1} C_{n + 1} s_{n} = 2^{n} S_{n}$

For any positive integer $m, n$ (with $n \geq m$ ), let $(m n) =^{n} C_{m}$ . Prove that $(m n) + (m n - 1) + (m n - 2) + .... + (m m) = (m + 1 n + 1)$ . Hence or otherwise, prove that $(m n) + 2 (m n - 1) + 3 (m n - 2) + .... + (n - m + 1) (m m) = (m + 2 n + 2)$ .

If $C_{r}$ stands for $^{n} C_{r}$ , then the sum of the series $\frac{2 ( \frac{n}{2} )! ( \frac{n}{2} )!}{n !} [C_{0}^{2} - 2 C_{1}^{2} + 3 C_{2}^{2} - .... + (- 1)^{n} (n + 1) C_{n}^{2}]$ , where $n$ is an even positive integer, is equal to

Suppose $det [\sum_{k = 0}^{n} k \sum_{k = 0}^{n}^{n} C_{k} k \sum_{k = 0}^{n}^{n} C_{k} k^{2} \sum_{k = 0}^{n}^{n} C_{k} 3^{k}] = 0$ , holds for some positive integer $n$ . Then $\sum_{k = 0}^{n} \frac{^{n} C _{k}}{k + 1}$ equals

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 $m \leq n$ .

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 $b_{n} =^{2 n + 1} C_{n + 1}$ .

Animated Solution for Mathematics - Binomial Theorem: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Prove that C0​−22C1​+32C2​−....+(−1)n(n+1)2Cn​=0,n>2, where Cr​=nCr​.