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

Hint: 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
\[\begin{align}
  & \Rightarrow 99=1+\left( n-1 \right)\times 2 \\
 & \Rightarrow 98=2\left( n-1 \right) \\
 & \Rightarrow n-1=49 \\
 & \Rightarrow n=50 \\
\end{align}\]
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.