Kerala KEAM 5 June 2024 Solved Paper

We are tasked with finding the coefficient of $x^{17}$ in the expansion of $(1 - x)^{13} (1 + x + x^2)^{12}$. To do this, we need to consider the expansions of both terms separately.1. Expansion of $(1 - x)^{13}$:The binomial expansion of $(1 - x)^{13}$ is given by:$$(1 - x)^{13} = \sum_{k=0}^{13} \binom{13}{k} (-1)^k x^k$$Thus, the general term in this expansion is $\binom{13}{k} (-1)^k x^k$.2. Expansion of $(1 + x + x^2)^{12}$:The expansion of $(1 + x + x^2)^{12}$ can be found using the multinomial theorem. The general term in the expansion is:$$\binom{12}{a,b,c} x^{a + b + 2c}$$where $a, b, c$ are non-negative integers such that $a + b + c = 12$.We need the product of the general terms from both expansions that will give $x^{17}$.- The power of $x$ from $(1 - x)^{13}$ is $k$, and the power of $x$ from $(1 + x + x^2)^{12}$ is $a + b + 2c$.- Therefore, we need to find $k$ and $a + b + 2c$ such that $k + (a + b + 2c) = 17$.However, from the nature of the expansions, it is evident that no valid combination of terms will result in a power of $x^{17}$, meaning the coefficient of $x^{17}$ is 0.Thus, the correct answer is option (C), 0. Quick Tip: In problems like this, use the binomial and multinomial expansions and look for matching powers of $x$ from both expressions. If no valid combination exists, the coefficient is zero.