Let m be the smallest positive integer such that the coefficient of x² in the expansion of (1 + x)²

Q: Why can't we just expand every term in the series? It seems like a lot of work.

Expanding every term is a trap! In JEE Advanced, time is your most precious resource. Recognizing the Hockey-stick identity transforms a 10-minute calculation into a 10-second realization. Always look for the pattern before you start the algebra.

Q: What is the Hockey-stick identity and why does it work?

Think of it as a committee selection problem. If you want to choose $k+1$ people from $n+1$ people, you can split the process based on who the $(k+1)$-th person is. It's a beautiful combinatorial argument that saves you from tedious summation.

Q: How do we know $n=5$ is the smallest?

We are looking for the smallest positive integer $n$. Since $m^2 = 51n + 1$, we simply test $n=1, 2, 3, 4, 5$. It's a finite search space. Don't overcomplicate it; sometimes the most direct path is the best one.

Question

Let $m$ be the smallest positive integer such that the coefficient of $x^2$ in the expansion of $(1 + x)^2 + (1 + x)^3 + ... + (1 + x)^{49} + (1 + mx)^{50}$ is $(3n + 1) ^{51}C_3$ for some positive integer $n$. Then the value of $n$ is .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The problem asks us to find the coefficient of $x^2$ in the expression:
$$(1 + x)^2 + (1 + x)^3 + \dots + (1 + x)^{49} + (1 + mx)^{50}$$

To simplify this, we split the expression into two distinct parts: the geometric-like series $\sum_{k=2}^{49} (1+x)^k$ and the final term $(1+mx)^{50}$. By isolating these, we transform a complex expansion into two manageable components.

### The Hockey-stick Magic

For any binomial expansion $(1 + x)^k$, the general term is $\binom{k}{r} x^r$. To find the coefficient of $x^2$, we set $r=2$, yielding the term $\binom{k}{2}$.

The coefficient of $x^2$ in the series is the sum:
$$\sum_{k=2}^{49} \binom{k}{2} = \binom{2}{2} + \binom{3}{2} + \binom{4}{2} + \dots + \binom{49}{2}$$

We invoke the **Hockey-stick Identity**, which states that $\sum_{r=k}^n \binom{r}{k} = \binom{n+1}{k+1}$. Applying this with $k=2$ and $n=49$, the entire sum collapses into:
$$\binom{50}{3}$$

### The Algebraic Dance

Next, we examine the term $(1+mx)^{50}$. The general term is $\binom{50}{r} (mx)^r$, and setting $r=2$ gives the coefficient:
$$\binom{50}{2} m^2$$

Combining these, the total coefficient of $x^2$ is $\binom{50}{3} + \binom{50}{2} m^2$. We are given that this must equal $(3n + 1) \binom{51}{3}$, leading to the master equation:
$$\binom{50}{3} + \binom{50}{2} m^2 = (3n + 1) \binom{51}{3}$$

### The Final Calculation

Expanding the combinations, we have:
$$\frac{50 \cdot 49 \cdot 48}{6} + \frac{50 \cdot 49}{2} m^2 = (3n + 1) \frac{51 \cdot 50 \cdot 49}{6}$$

Dividing the entire equation by the common factor $\frac{50 \cdot 49}{6}$, we simplify the expression significantly:
$$48 + 3m^2 = (3n + 1) \cdot 51$$

Expanding the right side yields $48 + 3m^2 = 153n + 51$. Subtracting $48$ from both sides and dividing by $3$, we arrive at the elegant relation:
$$m^2 = 51n + 1$$

We seek the smallest positive integer $n$ such that $51n + 1$ is a perfect square. Testing values for $n$, we find that for $n=5$:
$$m^2 = 51(5) + 1 = 255 + 1 = 256 = 16^2$$

Thus, the smallest positive integer $n$ is **5**.

Enter Numerical Value:

Visualized Solution

Understanding the Goal

Coefficient of $x^{2}$ in $(1 + x)^{k}$

The Hockey-stick Identity

Coefficient in $(1 + m x)^{50}$

Setting up the Equation

Expanding the Combinations

Simplifying to $m^{2} = 51 n + 1$

Finding the Smallest $n$

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Hockey-stick Magic

The Algebraic Dance

The Final Calculation

Similar Questions

If for some positive integer $n$ , the coefficients of three consecutive terms in the binomial expansion of $(1 + x)^{n + 5}$ are in the ratio $5 : 10 : 14$ , then the largest coefficient in this expansion is :

$r$ 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

The sum of the coefficients of three consecutive terms in the binomial expansion of $(1 + x)^{n + 2}$ , which are in the ratio $1 : 3 : 5$ , is equal to

If the sum of the coefficients of all even powers of $x$ in the product $(1 + x + x^{2} + ... + x^{2 n}) (1 - x + x^{2} - x^{3} + ... + x^{2 n})$ is 61, then $n$ is equal to . . . . .

Let $n$ be positive integer. If the coefficients of 2nd, 3rd, and 4th terms in the expansion of $(1 + x)^{n}$ are in A.P., then the value of $n$ is .........

For a positive integer $n$ , $(1 + \frac{1}{x})^{n}$ is expanded in increasing powers of $x$ . If three consecutive coefficients in this expansion are in the ratio, $2 : 5 : 12$ , then $n$ is equal to

The coefficients of three consecutive terms of $(1 + x)^{n + 5}$ are in the ratio $5 : 10 : 14$ . Then $n = .........$

The coefficient of $x^{48}$ in $(1 + x) + 2 (1 + x)^{2} + 3 (1 + x)^{3} + \dots + 100 (1 + x)^{100}$ is equal to

The sum, of the coefficients of the first 50 terms in the binomial expansion of $(1 - x)^{100}$ , is equal to

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:

Enter Numerical Value:

Visualized Solution

Understanding the Goal

Coefficient of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2 in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (1+x)k

The Hockey-stick Identity

Coefficient in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (1+mx)50

Setting up the Equation

Expanding the Combinations

Simplifying to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m2=51n+1

Finding the Smallest .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Hockey-stick Magic

The Algebraic Dance

The Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r and n are positive integers r>1,n>2 and coefficient of (r+2)th term and 3rth term in the expansion of (1+x)2n are equal, then n equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of the coefficients of three consecutive terms in the binomial expansion of (1+x)n+2, which are in the ratio 1:3:5, is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the sum of the coefficients of all even powers of x in the product (1+x+x2+...+x2n)(1−x+x2−x3+...+x2n) is 61, then n is equal to . . . . .

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let n be positive integer. If the coefficients of 2nd, 3rd, and 4th terms in the expansion of (1+x)n are in A.P., then the value of n is .........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For a positive integer n, (1+x1​)n is expanded in increasing powers of x. If three consecutive coefficients in this expansion are in the ratio, 2:5:12, then n is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The coefficients of three consecutive terms of (1+x)n+5 are in the ratio 5:10:14. Then n=.........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The coefficient of x48 in (1+x)+2(1+x)2+3(1+x)3+⋯+100(1+x)100 is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum, of the coefficients of the first 50 terms in the binomial expansion of (1−x)100, is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the coefficients of x4,x5 and x6 in the expansion of (1+x)n are in the arithmetic progression, then the maximum value of n is: