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$\left(x-\frac{3}{x^{2}}\right)^{m}, x \neq 0,$ जहाँ $m$ एक प्राकृत संख्या है, के प्रसार में पहले तीन पदों के गुणांकों का योग $559$ है। प्रसार में $x^{3}$ वाला पद ज्ञात कीजिए।
$ - 5940{x^3}$
$ - 5940{x^3}$
$ - 5940{x^3}$
$ - 5940{x^3}$
Solution
The coefficients of the first three terms of ${\left( {x – \frac{3}{{{x^2}}}} \right)^m}$ are $^m{C_0},( – 3){\,^m}{C_1}$ and $\,9{\,^m}{C_2}$. Therefore, by the given condition, we have
$^m{C_0} – 3{\,^m}{C_1} + 9{\,^m}{C_2} = 559,$ i.e., $1 – 3m + \frac{{9m(m – 1)}}{2} = 559$
which gives $m=12$ ( $m$ being a natural number).
Now ${T_{r + 1}} = {\,^{12}}{C_r}{x^{12 – r}}{\left( { – \frac{3}{{{x^2}}}} \right)^r} = {\,^{12}}{C_r}{( – 3)^r} \cdot {x^{12 – 3r}}$
Since we need the term containing $x^{3}$, so put $12-3 r=3$ i.e., $r=3$
Thus, the required term is ${\,^{12}}{C_3}{( – 3)^3}{x^3},$ i.e., $-5940 x^{3}$
