Question

The co-efficient of ${{x}^{49}}$ in the product $\left( x-1 \right)\left( x-3 \right)......\left( x-99 \right)$ is: -
(a) -99
(b) 1
(c) -2500
(d) None of these

Answer 1

First find the number of terms that are multiplied from (x - 1) to (x - 99) to determine the maximum power of x which we can get. Find all the possible methods to get ${{x}^{49}}$ by forming a general pattern of the coefficients of multiplied terms. Add all the coefficients to get the answer.

Complete step-by-step solution
Here, we have been provided with the expression: - $\left( x-1 \right)\left( x-3 \right)......\left( x-99 \right)$ and we have to find the co – efficient of ${{x}^{49}}$. First let us determine the number of terms.
Now, we can see that the constant term is starting from 1 and ending at 99 with a common difference of 2. So, apply the formula for the ${{n}^{th}}$ term of an A.P., we get,
$\Rightarrow {{T}_{n}}=a+\left( n-1 \right)d$
Here, a = first term = 1
d = common difference = 2
${{T}_{n}}$ = ${{n}^{th}}$ term = 99
n = number of terms

& \Rightarrow 99=1+\left( n-1 \right)\times 2 \\\ & \Rightarrow 98=2\left( n-1 \right) \\\ & \Rightarrow n-1=49 \\\ & \Rightarrow n=50 \\\ \end{aligned}$$ So, there are 50 terms in the given expression. Therefore, the maximum power of the variable x will be 50. Now, when will multiply the terms starting from (x - 1) and ending at (x - 97), i.e. 49 terms then we will get $${{x}^{49}}$$as the first term whose coefficient will be 1. $$\Rightarrow \left( x-1 \right)\left( x-3 \right)\left( x-5 \right)......\left( x-1 \right)={{x}^{49}}+M{{x}^{48}}+......$$ We have (x - 99) as the $${{50}^{th}}$$ term. So, to get $${{x}^{49}}$$ we have to multiply $${{x}^{49}}$$ with the constant term and $${{x}^{48}}$$ with x, in the $${{50}^{th}}$$ term, that is (x - 99). $$\Rightarrow $$ Co – efficient of $${{x}^{49}}$$ = -99 + M – (1) Let us try to find M now. Let us consider the following multiplication of terms: - (i) $(x - 1)$, here the co – efficient of $${{x}^{0}}$$ is -1 = $$-{{\left( 1 \right)}^{2}}$$. (ii) $(x - 1) (x - 3)$ = $${{x}^{2}}-4x+3$$, here the co – efficient of $${{x}^{1}}$$ is $$-4= -{{\left( 2 \right)}^{2}}$$. (iii) $(x - 1) (x - 3) (x - 5)$ = $${{x}^{3}}-9{{x}^{2}}+23x-15$$, here the co – efficient of $${{x}^{2}}$$ is $$-9= -{{\left( 3 \right)}^{2}}$$. So, on observing the pattern, we can conclude that when the product of first 49 terms takes place then the co – efficient of $${{x}^{48}}$$ will be $$-{{\left( 49 \right)}^{2}}$$. Therefore, when $$-{{\left( 49 \right)}^{2}}{{x}^{48}}$$ will be multiplied with x in the $${{50}^{th}}$$ term then we will get $${{x}^{49}}$$ whose co – efficient will be $$-{{\left( 49 \right)}^{2}}$$. Therefore, we have using equation (1): - Co-efficient of $${{x}^{49}}=-99+\left[ -{{\left( 49 \right)}^{2}} \right]$$ Co-efficient of $${{x}^{49}}=-99-2401=-2500$$ **Hence, the option (c) is the correct answer.** **Note:** One may note that it is not possible for us to multiply all the 50 terms and check the coefficients of $${{x}^{50}},{{x}^{49}},{{x}^{48}},....$$ and so on. So, we must find some general patterns to get the answer. First of all, you have to check how can we get $${{x}^{49}}$$ or whichever exponent of x is asked. Consider all the cases one by one and get the coefficients. At last, add all the coefficients with their signatures to get the answer.