First slide

Multinomial expansion

Question

Coefficient of x² in the expansion of ${(x^{3} + 2 x^{2} + x + 4)}^{15}$ is

Moderate

A

Prime
B

Composite
C

0
D

Perfect square

Solution

Let ${(x^{3} + 2 x^{2} + x + 4)}^{15} = a_{0} + a_{1} x + a_{2} x^{2} + \dots$
Differentiating w.r.t. x, we get
$\begin{matrix} 15 {(x^{3} + 2 x^{2} + x + 4)}^{14} (3 x^{2} + 4 x + 1) \\ = a_{1} + 2 a_{2} x + 3 a_{3} x^{2} + \dots \end{matrix}$
Again, differentiating w.r.t x, we get
$15 \{{(x^{3} + 2 x^{2} + x + 4)}^{14} (6 x + 4) + 14 {(x^{3} + 2 x^{2} + x + 4)}^{13} {(3 x^{2} + 4 x + 1)}^{2}\}$
$= 2 a_{2} + 6 a_{3} x + \dots$
Now, putting x = 0, we get
$\begin{array}{l} 15 (4 \times 4^{14} + 14 \times 4^{13}) = 2 a_{2} \\ \Rightarrow 2 a_{2} = 15 \times 4^{13} \times (16 + 14) \\ \Rightarrow a_{2} = (15)^{2} \times 2^{26} \end{array}$
which is a perfect square.

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