Suppose det [[Σ k, Σ ⁿCₖ k²],[Σ ⁿCₖ k, Σ ⁿCₖ 3ᵏ]] = 0, holds for some positive integer n. Then Σ (k=

Q: Wait, why do we rewrite $k^2$ as $k(k-1) + k$? Isn't that just making it more complicated?

That is a fantastic question! It feels like we are adding complexity, but we are actually performing a 'surgical' simplification. Binomial coefficients ${}^nC_k$ are defined by the factorial formula, and they play very nicely with terms like $k$ and $k(k-1)$ because these terms cancel out the $k!$ in the denominator. By splitting $k^2$ into $k(k-1) + k$, we transform a difficult sum into two standard, well-known binomial identities. It is a classic 'divide and conquer' strategy!

Q: Why can't we just expand the determinant directly without simplifying the elements first?

You absolutely could, but you would be walking into a massive algebraic trap! If you expand the determinant with the raw summation expressions, you will end up with a nightmare of nested summations and products that are nearly impossible to simplify. In JEE Advanced, the 'brute force' method is almost always a trap. The path to the solution lies in simplifying each element to its most elegant form first.

Q: Is there a faster way to calculate the final sum $\sum_{k=0}^{n} \frac{{}^nC_k}{k+1}$?

Yes, there is! You can use the powerful identity $\frac{{}^nC_k}{k+1} = \frac{{}^{n+1}C_{k+1}}{n+1}$. This identity is essentially the discrete version of integration. By using it, the sum transforms into a simple sum of binomial coefficients, which is just a power of $2$. It turns a potentially tedious calculation into a one-step simplification.

Question

Suppose $\det \begin{bmatrix} \sum_{k=0}^{n} k & \sum_{k=0}^{n} {}^nC_k k^2 \ \sum_{k=0}^{n} {}^nC_k k & \sum_{k=0}^{n} {}^nC_k 3^k \end{bmatrix} = 0$, holds for some positive integer $n$. Then $\sum_{k=0}^{n} \frac{{}^nC_k}{k+1}$ equals

Sigma Percentile · Accepted Answer

### The Matrix of Mystery

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dismantle a problem that, at first glance, looks like a fortress of complexity.

We are presented with a $2 	imes 2$ matrix, let's call it $A$, where the elements are summations of binomial coefficients. The problem states that the determinant of this matrix is zero, and from that, we must extract the value of $n$ and compute a final sum.

### Decoding the Elements

Our first phase is to decode the matrix. We have four elements, and we must treat each one as a mini-challenge.

The top-left element is $a_{11} = \sum_{k=0}^{n} k$. This is the sum of the first $n$ natural numbers:
$$a_{11} = \frac{n(n+1)}{2}$$

Next, the bottom-left element is $a_{21} = \sum_{k=0}^{n} {}^{n}C_{k} k$. Using the identity $k \cdot {}^{n}C_{k} = n \cdot {}^{n-1}C_{k-1}$, we pull the constant $n$ out of the summation to get the sum of all binomial coefficients for index $n-1$:
$$a_{21} = n \cdot 2^{n-1}$$

Now, the top-right element is $a_{12} = \sum_{k=0}^{n} {}^{n}C_{k} k^2$. We rewrite $k^2$ as $k(k-1) + k$ to split the sum into two parts: $\sum {}^{n}C_{k} k(k-1) + \sum {}^{n}C_{k} k$. 
$$a_{12} = n(n-1)2^{n-2} + n2^{n-1} = n(n+1)2^{n-2}$$

Finally, the bottom-right element is $a_{22} = \sum_{k=0}^{n} {}^{n}C_{k} 3^k$. By the Binomial Theorem $(1+x)^n = \sum_{k=0}^{n} {}^{n}C_{k} x^k$, setting $x=3$ yields:
$$a_{22} = (1+3)^n = 4^n$$

### The Determinant Dance

Now that we have our four elements, the matrix $A$ is defined as:

$$A = \begin{bmatrix} \frac{n(n+1)}{2} & n(n+1)2^{n-2} \ n2^{n-1} & 4^n \end{bmatrix}$$

We are given $\det(A) = 0$. Setting the product of the principal diagonal minus the product of the off-diagonal to zero:

$$\left( \frac{n(n+1)}{2} ight) (4^n) - \left( n(n+1)2^{n-2} ight) \left( n2^{n-1} ight) = 0$$

Simplifying the powers of $2$, where $4^n = 2^{2n}$ and $2^{n-2} \cdot 2^{n-1} = 2^{2n-3}$, the equation becomes:

$$\frac{n(n+1)}{2} 2^{2n} - n^2(n+1) 2^{2n-3} = 0$$

Factoring out $n(n+1)2^{2n-3}$ (which is non-zero for positive integers $n$), we obtain:

$$\frac{2^3}{2} - n = 0 \Rightarrow 4 - n = 0 \Rightarrow \mathbf{n = 4}$$

### The Final Flourish

With $n=4$, we evaluate $S = \sum_{k=0}^{4} \frac{{}^{4}C_{k}}{k+1}$. Using the identity $\frac{{}^{n}C_{k}}{k+1} = \frac{{}^{n+1}C_{k+1}}{n+1}$, we substitute $n=4$:

$$S = \frac{1}{5} \sum_{k=0}^{4} {}^{5}C_{k+1}$$

As $k$ ranges from $0$ to $4$, the index $k+1$ ranges from $1$ to $5$. The sum is ${}^{5}C_{1} + {}^{5}C_{2} + {}^{5}C_{3} + {}^{5}C_{4} + {}^{5}C_{5}$. Since the sum of all binomial coefficients for $n=5$ is $2^5 = 32$, and ${}^{5}C_{0} = 1$ is missing:

$$S = \frac{1}{5} (32 - 1) = \frac{31}{5} = \mathbf{6.2}$$

Animated Solution for Mathematics - Binomial Theorem: 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

Enter Numerical Value:

Visualized Solution

The Determinant Equation

Evaluating $a_{11}$

Evaluating $a_{21}$

Evaluating $a_{12}$

Evaluating $a_{22}$

Setting up the Determinant

Expanding the Determinant

Simplifying the Equation

Solving for $n$

The Target Expression

Applying Binomial Property

Evaluating the Sum

Final Calculation

The Sigma Insight: Properties of Binomial Coefficients

The Matrix of Mystery

Decoding the Elements

The Determinant Dance

The Final Flourish

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Enter Numerical Value:

Visualized Solution

The Determinant Equation

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