For some n ≠ 10, let the coefficients of the 5th, 6th and 7th terms in the binomial expansion of (1+

Q: Why can't I just expand the factorials immediately?

Expanding factorials early leads to a complex algebraic expression that is prone to errors. Using Pascal's Identity allows you to simplify the expression significantly before dealing with the factorials, making the algebra much cleaner and more elegant.

Q: How do I know which terms are the middle ones for the largest coefficient?

For a binomial expansion $(1+x)^m$, if $m$ is even, the largest coefficient is the middle term at $r = m/2$. If $m$ is odd, there are two middle terms at $r = (m-1)/2$ and $r = (m+1)/2$, and both have the same largest coefficient.

Q: Why is $n=10$ rejected?

The problem explicitly states $n \neq 10$. In competitive exams, always check the constraints provided in the question, as they often filter out extraneous solutions that arise from the algebraic manipulation.

Question

For some $n
e10$, let the coefficients of the 5 th, 6 th and 7 th terms in the binomial expansion of $(1+x)^{n+4}$ be in A.P. Then the largest coefficient in the expansion of $(1+x)^{n+4}$ is:

Options:

(A) 20

(B) 10

(C) 35

(D) 70

Sigma Percentile · Accepted Answer

### The Elegance of Binomial Symmetry

Welcome, future engineers! Today, we are diving into a problem that might seem like a dry algebraic exercise at first glance, but it is actually a beautiful dance of symmetry and logic.

We are dealing with the binomial expansion of $(1+x)^{n+4}$. Our mission is to find the largest coefficient, but only after we decode a hidden constraint about the 5th, 6th, and 7th terms.

### Phase 1: Setting the Stage

First, let's orient ourselves. The general term of a binomial expansion $(1+x)^m$ is given by $T_{r+1} = {}^{m}C_r \cdot x^r$.

In our case, the power is $m = n+4$. So, the coefficients of the 5th, 6th, and 7th terms correspond to $r=4$, $r=5$, and $r=6$, respectively.

These are ${}^{n+4}C_4$, ${}^{n+4}C_5$, and ${}^{n+4}C_6$. This is our starting point.

### Phase 2: The A.P. Condition

We are told these three coefficients are in an Arithmetic Progression (A.P.). This means the middle term is the average of the other two, or more simply, $2 \cdot 	ext{middle} = 	ext{first} + 	ext{third}$.

Mathematically, this gives us the equation:

$$2({}^{n+4}C_5) = {}^{n+4}C_4 + {}^{n+4}C_6$$

Most students would immediately reach for the factorial formula $\frac{n!}{r!(n-r)!}$ and start expanding. I urge you: stop! That path leads to a forest of factorials that is very easy to get lost in.

### Phase 3: The Elegant Shortcut

This is where we use the power of Pascal's Identity: ${}^{m}C_r + {}^{m}C_{r-1} = {}^{m+1}C_r$. We want to force this identity to appear.

If we add $2({}^{n+4}C_5)$ to both sides of our A.P. equation, we get:

$$4({}^{n+4}C_5) = ({}^{n+4}C_4 + {}^{n+4}C_5) + ({}^{n+4}C_5 + {}^{n+4}C_6)$$

Look at the magic! The terms in the parentheses are now in the perfect form for Pascal's Identity. The first pair becomes ${}^{n+5}C_5$, and the second pair becomes ${}^{n+5}C_6$.

Our equation now looks like this:

$$4({}^{n+4}C_5) = {}^{n+5}C_5 + {}^{n+5}C_6$$

And we can apply Pascal's Identity one more time to the right side to get ${}^{n+6}C_6$. Now, the equation is much simpler:

$$4({}^{n+4}C_5) = {}^{n+6}C_6$$

### Phase 4: Solving for $n$

Now, we can safely expand the factorials. With the equation:

$$4 \cdot \frac{(n+4)!}{5!(n-1)!} = \frac{(n+6)!}{6!n!}$$

We can cancel terms. After simplifying, we arrive at the quadratic equation $n^2 - 13n + 30 = 0$.

Factoring this gives $(n-10)(n-3) = 0$. Since the problem explicitly states $n 
eq 10$, we must have $n=3$.

### Phase 5: The Grand Finale

With $n=3$, our expansion becomes $(1+x)^{3+4} = (1+x)^7$. We need the largest coefficient.

For an odd power like 7, the largest coefficients are the middle ones, at $r=3$ and $r=4$. Both are equal to ${}^{7}C_3$.

Calculating this, we get:

$$\frac{7 	imes 6 	imes 5}{3 	imes 2 	imes 1} = 35$$

And there it is—the **answer is 35**. It wasn't just about the calculation; it was about choosing the path of least resistance. Keep this mindset, and you will conquer any JEE problem!

Animated Solution for Mathematics - Binomial Theorem: For some $n \neq = 10$ , let the coefficients of the 5 th, 6 th and 7 th terms in the binomial expansion of $(1 + x)^{n + 4}$ be in A.P. Then the largest coefficient in the expansion of $(1 + x)^{n + 4}$ is:

Select Answer:

Visualized Solution

Identify the Binomial Expression

Extract the Coefficients

Apply the A.P. Condition

Algebraic Manipulation Trick

Simplify using Pascal's Identity

Apply Pascal's Identity

Expand the Combinations

Simplify the Factorials

Form the Quadratic Equation

Solve for $n$

Find the Expansion Power

Determine the Largest Coefficient

Final Conclusion

The Sigma Insight: Properties of Binomial Coefficients

The Elegance of Binomial Symmetry

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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