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The coefficient of $x^3$ in the expansion of (1 + 2$x$ + 3$x^2$)$^6$ is equal to twice the coefficient of $x^4$ in the expansion of $(1 - a x^2)^5$.

Find all possible value of constant $a$.

I am actually getting $a=8$ or $a=-8$ using the binomial theorem or Pascal's Triangle.

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Hint $:$ Observe that $$\begin{align} (1+2x+3x^2)^6 & = 1 + \binom 6 1 (2x+3x^2) + \binom 6 2 (2x+3x^2)^2 + \binom 6 3 (2x+3x^2)^3 + \cdots + (2x+3x^2)^6. \\ & = 1 + \binom 6 1 x(2+3x) + \binom 6 2 x^2(2+3x)^2 + \binom 6 3 x^3(2+3x)^3 + \cdots + x^6(2+3x)^6. \end{align}$$ Can you see now?

So the coefficient of $x^3$ in $(1+2x+3x^2)^6$ is $\left (12 \times \binom 6 2 \right ) + \left (8 \times \binom 6 3 \right ) = 180 + 160 = 340.$

Now the coefficient of $x^4$ in $(1-ax^2)^5$ is $\binom 5 2 a^2 = 10 a^2.$

So according to the given problem we must have $2 \times 10a^2 = 340 \implies a^2 = 17.$ Therefore $a = \pm {\sqrt {17}}.$

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  • $\begingroup$ Thanks a lot Sir. I use a different approach but I did not get the above answer. I have understood the logic. Thanks a lot. $\endgroup$ – Sidoine Samo Djossi Apr 4 '19 at 19:42
  • $\begingroup$ You are most welcome @Sidoine Samo Djossi. $\endgroup$ – little o Apr 4 '19 at 19:46

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