In the expansion of \[(x - 1)(x - 2)(x - 3)...(x - 18)\] , the coefficient of ${x^{17}}$ is
$(a){\text{ }}684$
$(b){\text{ }} - 171$
$(c){\text{ }}171$
$(a){\text{ - 342}}$
Last updated date: 25th Mar 2023
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Answer
308.4k+ views
(Hint: The coefficient of ${x^{17}}$ is calculated by the addition of the given series. This can be understood as:-If $(x - 1)(x - 2) = {x^2} - 3x + 2$then coefficient of $x = -1 +(- 2) = -3$.
In the question, we are given the expansion as
\[(x - 1)(x - 2)(x - 3)...(x - 18)\]
Here, we can have the maximum power of $x = 18$
Now, in order to find out the coefficient of ${x^{17}}$
We will add the coefficients of the given expansion
Such that,
\[ = - 1 + ( - 2) + ( - 3) + ...( - 18)\]
\[ = - 1 - 2 - 3... - 18\]
\[ = - (1 + 2 + 3... + 18)\]
Now, we know that the sum of $n$ terms is equal to $\dfrac{{n(n + 1)}}{2}$
Here, we have $n = 18$
Therefore, we get the sum of these $18$ terms as
\[ = - \dfrac{{18(18 + 1)}}{2}\]
\[ = - \dfrac{{18(19)}}{2}\]
\[ = - 9(19)\]
\[ = - 171\]
Which is the required coefficient of the ${x^{17}}$
Therefore, the required solution is $(b){\text{ - 171}}$.
Note: In order to solve these types of questions, the students must have an adequate knowledge of the expansion of the series.
In the question, we are given the expansion as
\[(x - 1)(x - 2)(x - 3)...(x - 18)\]
Here, we can have the maximum power of $x = 18$
Now, in order to find out the coefficient of ${x^{17}}$
We will add the coefficients of the given expansion
Such that,
\[ = - 1 + ( - 2) + ( - 3) + ...( - 18)\]
\[ = - 1 - 2 - 3... - 18\]
\[ = - (1 + 2 + 3... + 18)\]
Now, we know that the sum of $n$ terms is equal to $\dfrac{{n(n + 1)}}{2}$
Here, we have $n = 18$
Therefore, we get the sum of these $18$ terms as
\[ = - \dfrac{{18(18 + 1)}}{2}\]
\[ = - \dfrac{{18(19)}}{2}\]
\[ = - 9(19)\]
\[ = - 171\]
Which is the required coefficient of the ${x^{17}}$
Therefore, the required solution is $(b){\text{ - 171}}$.
Note: In order to solve these types of questions, the students must have an adequate knowledge of the expansion of the series.
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