First slide

Binomial theorem for positive integral Index

Question

If the coefficients of x² and x³ are both zero, in the expansion of the expression $(1 + ax + {bx}^{2}) (1 - 3 x)^{15}$ in powers of x then the ordered pair
(a, b) is equal to

Moderate

A

(28,315)
B

(-21,714)
C

(28,861)
D

(-54,315)

Solution

Given expression is $(1 + ax + {bx}^{2}) (1 - 3 x)^{15}$ in the expansion of binomial $(1 - 3 x)^{15} the (r + 1)$ th term is
$T_{r + 1} =^{15} C_{r} (- 3 x)^{r} =^{15} C_{r} (- 3)^{r} x^{r}$
Now, coefficient of x², in the expansion of
$(1 + ax + {bx}^{2}) (1 - 3 x)^{15}$
$^{15} C_{2} (- 3)^{2} + a^{15} C_{1} (- 3)^{1} + b^{15} C_{0} (- 3)^{0} = 0$ (given)
$\Rightarrow (105 \times 9) - 45 a + b = 0 \Rightarrow 45 a - b = 945$ …..(1)
similarly, the" coefficient of x³, in the expansion of
$(1 + ax + {bx}^{2}) (1 - 3 x)^{15} is$
$^{15} C_{3} (- 3)^{3} + a^{15} C_{2} (- 3)^{2} + b^{15} C_{1} (- 3)^{1} = 0$
$\begin{array}{lrl} \Rightarrow - 12285 + 945 a - 45 b = 0 \\ \Rightarrow 63 a - 3 b = 819 \Rightarrow 21 a - b = 273 - - - - (i) \end{array}$
From Eqs. (i) and (ii), we get
$\begin{array}{l} 24 a = 672 \Rightarrow a = 28 \\ so, b = 315 \Rightarrow (a, b) = (28, 315) \end{array}$

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