Sum of coefficients of first three terms in expansion of left(x-frac{3}{x^{2}}right)^{m}, x ≠ 0, m

Hindi

7.Binomial Theorem

hard

The sum of the coefficients of the first three terms in the expansion of $\left(x-\frac{3}{x^{2}}\right)^{m}, x \neq 0, m$ being a natural number, is $559 .$ Find the term of the expansion containing $x^{3}$

$ - 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., $-5940 x^{3}$

Standard 11

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Solution

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