Name: Finite Series in Binomial Coefficients
Uploaded: 2019-07-19
Duration: 6 min 12 s
Description: Learn the concept of summation of finite series in binomial coefficients and explore the use of derivative to find the sum of series as well by watching this...

Question

Find the coefficient of:

(a) $x^{6}$ in the expansion of ${(a x^{2} + b x + c)}^{9}$

(b) $x^{2} y^{3} z^{4}$ in the expansion of ${(a x - b y + c z)}^{9}$ .

(c) $a^{2} b^{3} c^{4} d$ in the expansion of ${(a - b - c + d)}^{10}$ .

Accepted Answer

$(i)$ Using Multinomial theorem, Coefficient of $x^{6}$ in the expansion of ${(a x^{2} + b x + c)}^{9}$ is given by,

$\frac{9!}{(r_{1}!) (r_{2}!) (r_{3}!)} {(a x^{2})}^{r_{1}} {(b x)}^{r_{2}} {(c)}^{r_{3}}$

or, $\frac{9!}{(r_{1}!) (r_{2}!) (r_{3}!)} a^{r_{1}} b^{r_{2}} c^{r_{3}} x^{2 r_{1} + r_{2}}$

Clearly, $2 r_{1} + r_{2} = 6 & r_{1} + r_{2} + r_{3} = 9$ which is possible if and only if $(r_{1}, r_{2}) = (0, 6), (1, 4), (2, 2), (3, 0)$

Hence required coefficient is $\frac{9!}{(0!) (6!) (3!)} a^{0} b^{6} c^{3} + \frac{9!}{(1!) (4!) (4!)} a^{1} b^{4} c^{4} + \frac{9!}{(2!) (2!) (5!)} a^{2} b^{2} c^{5} + \frac{9!}{(3!) (0!) (6!)} a^{3} b^{0} c^{6}$

or, $\frac{9!}{(6!) (3!)} b^{6} c^{3} + \frac{9!}{(4!) (4!)} a b^{4} c^{4} + \frac{9!}{(2!) (2!) (5!)} a^{2} b^{2} c^{5} + \frac{9!}{(3!) (6!)} a^{3} c^{6}$

This is the final simplified result.

$(i i)$ Using Multinomial theorem, Coefficient of $x^{2} y^{3} z^{4}$ in the expansion of ${(a x - b y + c z)}^{9}$ is given by,

$\frac{9!}{(r_{1}!) (r_{2}!) (r_{3}!)} {(a x)}^{r_{1}} {(- b y)}^{r_{2}} {(c z)}^{r_{3}}$

or, $\frac{{(- 1)}^{r_{2}} 9!}{(r_{1}!) (r_{2}!) (r_{3}!)} a^{r_{1}} b^{r_{2}} c^{r_{3}} x^{r_{1}} y^{r_{2}} z^{r_{3}}$

Clearly, here $(r_{1}, r_{2}, r_{3}) = (2, 3, 4)$

Hence required coefficient is $\frac{{(- 1)}^{3} 9!}{(2!) (3!) (4!)} a^{2} b^{3} c^{4} = - 1260 (a^{2} b^{3} c^{4})$

This is the final simplified result.

$(i i i)$ Using Multinomial theorem, Coefficient of $a^{2} b^{3} c^{4} d$ in the expansion of ${(a - b + c - d)}^{10}$ is given by,

$\frac{10!}{(r_{1}!) (r_{2}!) (r_{3}!) (r_{4}!)} {(a)}^{r_{1}} {(- b)}^{r_{2}} {(c)}^{r_{3}} {(- d)}^{r_{4}}$

or, $\frac{{(- 1)}^{r_{2} + r_{4}} 10!}{(r_{1}!) (r_{2}!) (r_{3}!) (r_{4}!)} a^{r_{1}} b^{r_{2}} c^{r_{3}} d^{r_{4}}$

Clearly, here $(r_{1}, r_{2}, r_{3}, r_{4}) = (2, 3, 4, 1)$

Hence required coefficient is $\frac{{(- 1)}^{3 + 1} 10!}{(2!) (3!) (4!) (1!)} a^{2} b^{3} c^{4} d^{1} = 12600 (a^{2} b^{3} c^{4} d)$

This is the final simplified result.

Find the coefficient of:

(a) $x^{6}$ in the expansion of ${(a x^{2} + b x + c)}^{9}$

(b) $x^{2} y^{3} z^{4}$ in the expansion of ${(a x - b y + c z)}^{9}$ .

(c) $a^{2} b^{3} c^{4} d$ in the expansion of ${(a - b - c + d)}^{10}$ .

Important Questions on Binomial Theorem

Learn and Prepare for any exam you want !

Find the coefficient of: (a) x6 in the expansion of ax2+bx+c9 (b) x2y3z4 in the expansion of ax-by+cz9. (c) a2b3c4d in the expansion of a-b-c+d10.

Important Questions on Binomial Theorem

Learn and Prepare for any exam you want !

Find the coefficient of:

(a) $x^{6}$ in the expansion of ${(a x^{2} + b x + c)}^{9}$

(b) $x^{2} y^{3} z^{4}$ in the expansion of ${(a x - b y + c z)}^{9}$ .

(c) $a^{2} b^{3} c^{4} d$ in the expansion of ${(a - b - c + d)}^{10}$ .