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प्रश्न
If \[\frac{d}{dx}\left[ x^n - a_1 x^{n - 1} + a_2 x^{n - 2} + . . . + \left( - 1 \right)^n a_n \right] e^x = x^n e^x\] then the value of ar, 0 < r ≤ n, is equal to
पर्याय
\[\frac{n!}{r!}\]
\[\frac{\left( n - r \right)!}{r!}\]
\[\frac{n!}{\left( n - r \right)!}\]
none of these
उत्तर
(c) \[\frac{n!}{\left( n - r \right)!}\]
According to the given equation,
\[\frac{d}{dx}\left[ x^n - a_1 x^{n - 1} + a_2 x^{n - 2} + . . . + \left( - 1 \right)^n a_n \right] e^x = x^n e^x \]
\[ \Rightarrow \frac{d}{dx}\left[ x^n - a_1 x^{n - 1} + a_2 x^{n - 2} + . . . + \left( - 1 \right)^n a_n \right] e^x = \frac{d}{dx}\left[ x^n - n x^{n - 1} + n\left( n - 1 \right) x^{n - 2} + . . . + \left( - 1 \right)^n a_n \right] e^x \]
\[\text { Comparing the coefficients of the above equation we get }, \]
\[ a_1 = n\]
\[ a_2 = n\left( n - 1 \right)\]
\[\text { Similarly }, \]
\[ a_r = n\left( n - 1 \right)\left( n - 2 \right)\left( n - 3 \right) . . . \left( n - r + 1 \right)\]
\[ \Rightarrow a_r = \frac{n!}{\left( n - r \right)!}\]
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