JEE Mathematics 2024 Question with Solution

Given: The four consecutive coefficients in the expansion of $(1+x)^n$ are $2-p,\ p,\ 2-\alpha,\ \alpha$.

Find: The value of $p^2-\alpha^2+6\alpha+2p$.

Let these coefficients be

$$\binom{n}{k}=2-p, \quad \binom{n}{k+1}=p, \quad \binom{n}{k+2}=2-\alpha, \quad \binom{n}{k+3}=\alpha$$

Using Pascal's identity on the first two coefficients,

$$\binom{n}{k}+\binom{n}{k+1}=\binom{n+1}{k+1}=2$$

Similarly, for the last two coefficients,

$$\binom{n}{k+2}+\binom{n}{k+3}=\binom{n+1}{k+3}=2$$

So,

$$\binom{n+1}{k+1}=\binom{n+1}{k+3}$$

Since these are different positions in the same row, symmetry gives

$$(k+1)+(k+3)=n+1$$

Hence,

$$n=2k+3$$

Now compare the middle two coefficients:

$$\binom{n}{k+2}=\binom{2k+3}{k+2}=\binom{2k+3}{k+1}=\binom{n}{k+1}$$

Therefore,

$$p=2-\alpha$$

So,

$$p+\alpha=2$$

Now evaluate the expression:

$$p^2-\alpha^2+6\alpha+2p=(p-\alpha)(p+\alpha)+6\alpha+2p$$

Using $p+\alpha=2$,

$$=(p-\alpha)\cdot 2+6\alpha+2p=2p-2\alpha+6\alpha+2p=4p+4\alpha$$

Thus,

$$4p+4\alpha=4(p+\alpha)=4\cdot 2=8$$

Therefore, the value of the expression is $8$. Hence, the correct option is C.

The answer key marks option B, but the extracted solution working concludes $8$, so the solution is taken as authoritative.

Given: The consecutive coefficients are $2-p,\ p,\ 2-\alpha,\ \alpha$.

Find: $p^2-\alpha^2+6\alpha+2p$.

Add the first two and the last two coefficients:

$$(2-p)+p=2, \qquad (2-\alpha)+\alpha=2$$

So two binomial coefficients in the next row are equal. By symmetry of binomial coefficients, the middle two coefficients in the original row become equal, giving

$$p=2-\alpha \Rightarrow p+\alpha=2$$

Now write

$$p^2-\alpha^2+6\alpha+2p=(p-\alpha)(p+\alpha)+6\alpha+2p$$

Substitute $p+\alpha=2$:

$$=2(p-\alpha)+6\alpha+2p=4p+4\alpha=4(p+\alpha)=8$$

Therefore, the value is $8$ and the correct option is C.