Question

Let $(1 - x + x^{2})^{30} = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + ........ + a_{60}x^{60}$. Match List-I with List-Il and select the correct answer using the code given below the list.

LIST-I	LIST-II
(P) $\left(\frac{a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - a_{3}^{2} + ......}{a_{30}}\right)$	(1) 0
(Q) $4\left(\frac{a_{0} + a_{2} + a_{4} + ......}{(3^{30} + 1)}\right)$	(2) 1
(R) $6\left(\frac{a_{1} + a_{3} + a_{5} + ......}{(1 - 3^{30})}\right)$	(3) 2
(S) $a_{0}a_{1} - a_{1}a_{2} + ........$	(4) 3
	(5) 4

• A. (1) 0
• B. (2) 1
• C. (3) 2
• D. (4) 3
• E. (5) 4

Answer 1

P-2, Q-3, R-4, S-1

Answer 2

Let the given polynomial be $P(x) = (1 - x + x^{2})^{30} = a_{0} + a_{1}x + a_{2}x^{2} + \dots + a_{60}x^{60}$.

We evaluate $P(x)$ at specific values:

1. For $x=1$: $P(1) = (1 - 1 + 1^2)^{30} = 1^{30} = 1$. Also, $P(1) = a_0 + a_1 + a_2 + \dots + a_{60}$. So, $a_0 + a_1 + a_2 + \dots + a_{60} = 1$. (Equation 1)

2. For $x=-1$: $P(-1) = (1 - (-1) + (-1)^2)^{30} = (1 + 1 + 1)^{30} = 3^{30}$. Also, $P(-1) = a_0 - a_1 + a_2 - \dots + a_{60}$. So, $a_0 - a_1 + a_2 - \dots + a_{60} = 3^{30}$. (Equation 2)

We also know that $a_0 = P(0) = (1 - 0 + 0)^{30} = 1$.

Let's analyze each part of List-I:

Part (P): $\left(\frac{a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - a_{3}^{2} + ......}{a_{30}}\right)$

The sum in the numerator is $S = \sum_{k=0}^{60} (-1)^k a_k^2$. This sum is the coefficient of $x^0$ in the product $P(x) \cdot P(-1/x)$. $P(x) = (1 - x + x^2)^{30}$. $P(-1/x) = \left(1 - \left(-\frac{1}{x}\right) + \left(-\frac{1}{x}\right)^2\right)^{30} = \left(1 + \frac{1}{x} + \frac{1}{x^2}\right)^{30} = \left(\frac{x^2 + x + 1}{x^2}\right)^{30} = \frac{(1 + x + x^2)^{30}}{x^{60}}$.

So, $P(x) \cdot P(-1/x) = (1 - x + x^2)^{30} \cdot \frac{(1 + x + x^2)^{30}}{x^{60}}$ $= \frac{((1 - x + x^2)(1 + x + x^2))^{30}}{x^{60}}$ $= \frac{((1 + x^2)^2 - x^2)^{30}}{x^{60}}$ $= \frac{(1 + 2x^2 + x^4 - x^2)^{30}}{x^{60}}$ $= \frac{(1 + x^2 + x^4)^{30}}{x^{60}}$.

The coefficient of $x^0$ in this expression is the coefficient of $x^{60}$ in $(1 + x^2 + x^4)^{30}$. Let $y = x^2$. We need the coefficient of $y^{30}$ in $(1 + y + y^2)^{30}$. Let $(1 + y + y^2)^{30} = \sum_{k=0}^{60} A_k y^k$. Then the coefficient of $y^{30}$ is $A_{30}$. So, $S = A_{30}$.

Now we need to find $a_{30}$. Consider the relationship between the coefficients of $(1 - x + x^2)^n$ and $(1 + x + x^2)^n$. Let $(1 + x + x^2)^{30} = A_0 + A_1 x + A_2 x^2 + \dots + A_{60} x^{60}$. If we replace $x$ with $-x$ in the expansion of $(1 + x + x^2)^{30}$: $(1 - x + x^2)^{30} = A_0 - A_1 x + A_2 x^2 - \dots + A_{60} x^{60}$. Comparing this with $a_0 + a_1 x + a_2 x^2 + \dots + a_{60} x^{60}$, we get $a_k = (-1)^k A_k$.

Therefore, $a_{30} = (-1)^{30} A_{30} = A_{30}$. So, $\frac{S}{a_{30}} = \frac{A_{30}}{A_{30}} = 1$. Thus, (P) matches with (2).

Part (Q): $4\left(\frac{a_{0} + a_{2} + a_{4} + ......}{(3^{30} + 1)}\right)$

From Equation 1 and Equation 2: $(a_0 + a_1 + a_2 + \dots + a_{60}) + (a_0 - a_1 + a_2 - \dots + a_{60}) = 1 + 3^{30}$. $2(a_0 + a_2 + a_4 + \dots + a_{60}) = 1 + 3^{30}$. So, $a_0 + a_2 + a_4 + \dots + a_{60} = \frac{1 + 3^{30}}{2}$. Substitute this into the expression: $4\left(\frac{\frac{1 + 3^{30}}{2}}{3^{30} + 1}\right) = 4\left(\frac{1 + 3^{30}}{2(3^{30} + 1)}\right) = 4 \times \frac{1}{2} = 2$. Thus, (Q) matches with (3).

Part (R): $6\left(\frac{a_{1} + a_{3} + a_{5} + ......}{(1 - 3^{30})}\right)$

From Equation 1 and Equation 2: $(a_0 + a_1 + a_2 + \dots + a_{60}) - (a_0 - a_1 + a_2 - \dots + a_{60}) = 1 - 3^{30}$. $2(a_1 + a_3 + a_5 + \dots + a_{59}) = 1 - 3^{30}$. So, $a_1 + a_3 + a_5 + \dots + a_{59} = \frac{1 - 3^{30}}{2}$. Substitute this into the expression: $6\left(\frac{\frac{1 - 3^{30}}{2}}{1 - 3^{30}}\right) = 6\left(\frac{1 - 3^{30}}{2(1 - 3^{30})}\right) = 6 \times \frac{1}{2} = 3$. Thus, (R) matches with (4).

Part (S): $a_{0}a_{1} - a_{1}a_{2} + ........$

The sum is $S' = a_0 a_1 - a_1 a_2 + a_2 a_3 - a_3 a_4 + \dots - a_{59} a_{60}$. This can be written as $\sum_{k=0}^{59} (-1)^k a_k a_{k+1}$. We use the relation $a_k = (-1)^k A_k$. $S' = \sum_{k=0}^{59} (-1)^k ((-1)^k A_k) ((-1)^{k+1} A_{k+1})$ $S' = \sum_{k=0}^{59} (-1)^k (-1)^k (-1)^{k+1} A_k A_{k+1}$ $S' = \sum_{k=0}^{59} (-1)^{3k+1} A_k A_{k+1}$. If $k$ is even, $3k+1$ is odd. $(-1)^{3k+1} = -1$. If $k$ is odd, $3k+1$ is even. $(-1)^{3k+1} = 1$. So, $S' = -A_0 A_1 + A_1 A_2 - A_2 A_3 + A_3 A_4 - \dots + A_{59} A_{60}$.

Consider the coefficient of $x^{-1}$ in the expansion of $P(x) \cdot P'(1/x)$. This is not straightforward. Let's consider the derivative of $P(x)P(-x) = (1+x^2+x^4)^{30}$. $P(x)P(-x) = (1+x^2+x^4)^{30}$. Since $(1+x^2+x^4)^{30}$ is an even function (only contains even powers of $x$), its derivative must be an odd function (only contains odd powers of $x$). Let $Q(x) = (1+x^2+x^4)^{30}$. Then $Q'(x)$ has only odd powers of $x$. $P(x) = \sum a_k x^k$. $P'(x) = \sum k a_k x^{k-1}$. $P(-x) = \sum a_k (-1)^k x^k$. $P'(-x) = \sum k a_k (-1)^k (-x)^{k-1} (-1) = \sum k a_k (-1)^{k+1} x^{k-1}$. Derivative of $P(x)P(-x)$ is $P'(x)P(-x) + P(x)P'(-x)$. The coefficient of $x^0$ in this derivative would be 0.

Let's use the property $a_k = a_{60-k}$. This is true for symmetric polynomials. For $(1-x+x^2)^{30}$, the coefficients are symmetric, $a_k = a_{60-k}$. So $a_0 = a_{60}$, $a_1 = a_{59}$, $a_2 = a_{58}$, and so on. The sum $S' = a_0 a_1 - a_1 a_2 + a_2 a_3 - \dots - a_{59} a_{60}$. Using $a_{60} = a_0$, $a_{59} = a_1$, etc. The last term is $-a_{59}a_{60} = -a_1 a_0$. The sum becomes $a_0 a_1 - a_1 a_2 + a_2 a_3 - \dots - a_1 a_0$. If the number of terms is even, the sum will be 0. The terms are $a_k a_{k+1}$ for $k=0, \dots, 59$. There are 60 terms. $S' = (a_0 a_1 - a_{59} a_{60}) + (-a_1 a_2 + a_{58} a_{59}) + \dots$ Since $a_{59} = a_1$ and $a_{60} = a_0$, the last term $-a_{59}a_{60} = -a_1 a_0$. The first term is $a_0 a_1$. So $a_0 a_1 - a_1 a_0 = 0$. The second term is $-a_1 a_2$. The second to last term is $-a_{58}a_{59}$. $a_{58}=a_2$, $a_{59}=a_1$. So $-a_2 a_1$. This pattern seems to cancel out. Let's write it out: $S' = a_0 a_1 - a_1 a_2 + a_2 a_3 - a_3 a_4 + \dots + a_{58} a_{59} - a_{59} a_{60}$. Using $a_k = a_{60-k}$: $a_{59} = a_1$, $a_{60} = a_0$. So $-a_{59}a_{60} = -a_1 a_0$. $a_{58} = a_2$, $a_{59} = a_1$. So $a_{58}a_{59} = a_2 a_1$. $a_{57} = a_3$, $a_{58} = a_2$. So $-a_{57}a_{58} = -a_3 a_2$. The sum is: $(a_0 a_1 - a_1 a_0) + (-a_1 a_2 + a_1 a_2) + (a_2 a_3 - a_2 a_3) + \dots$ All terms cancel out in pairs. Thus, $S' = 0$. So, (S) matches with (1).

Summary of matches: (P) $\to$ (2) (Q) $\to$ (3) (R) $\to$ (4) (S) $\to$ (1)