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# Binomial Theorem

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Find the constant term in the expansion of (2z - 1/sqrt(z) + z^3)^9.

May 8, 2022

#4
+9462
+2

The complexity of this problem is not the same level as a homework problem, so I will work it out in full details.

Using trinomial theorem (which is just a generalization of binomial theorem), we have

$$\left(2z - \dfrac1{\sqrt z} + z^3\right)^9 = \displaystyle \sum_{\substack{a + b + c = 9\\a,b,c\text{ nonnegative integers}}} \binom{9}{a,b,c} (2z)^a \left(-\dfrac1{\sqrt z}\right)^b (z^3)^c$$

where $$\displaystyle\binom{9}{a,b,c}$$ denotes the trinomial coefficient $$\displaystyle\binom{n}{a,b,c} = \dfrac{n!}{a! b! c!}$$.

So basically, we first have to find nonnegative integers a, b with a + b <= 9 where a - b/2 + 3(9 - a - b) = 0.

If we move b/2 to the other side of the equation, we get b/2 = a + 3(9 - a - b). Since the right-hand side is an integer, the left-hand side, i.e., b/2, must also be an integer. So we can brute-force and try b = 0, 2, 4, 6, 8 and see which one works. Trying one by one gives a = 10, b = 2 or a = 3, b = 6 as the only solutions with b in the range 0 <= b <= 9. Since a + b <= 9, we can reject the first solution. Doing so, we have $$\begin{cases}a = 3\\b = 6\end{cases}$$ as the only nonnegative integer solution to the equation.

Now, the constant term is $$\displaystyle \binom{9}{3,6,0} (2z)^3 \left(-\dfrac1{\sqrt z}\right)^6(z^3)^0 = \binom{9}3 2^3 = 672$$.

May 8, 2022

#1
+14406
+1

The constant term is 672.

May 8, 2022
#3
+14406
+1

I googled the result so the responder can compare their result. I wish him fun and patience with this incredibly complicated task!

!

asinus  May 8, 2022
#2
+118448
+1

(2z - 1/sqrt(z) + z^3)^9.

$$(2z - \frac{1}{\sqrt z}+ z^3)^9$$

May 8, 2022
#4
+9462
+2

The complexity of this problem is not the same level as a homework problem, so I will work it out in full details.

Using trinomial theorem (which is just a generalization of binomial theorem), we have

$$\left(2z - \dfrac1{\sqrt z} + z^3\right)^9 = \displaystyle \sum_{\substack{a + b + c = 9\\a,b,c\text{ nonnegative integers}}} \binom{9}{a,b,c} (2z)^a \left(-\dfrac1{\sqrt z}\right)^b (z^3)^c$$

where $$\displaystyle\binom{9}{a,b,c}$$ denotes the trinomial coefficient $$\displaystyle\binom{n}{a,b,c} = \dfrac{n!}{a! b! c!}$$.

So basically, we first have to find nonnegative integers a, b with a + b <= 9 where a - b/2 + 3(9 - a - b) = 0.

If we move b/2 to the other side of the equation, we get b/2 = a + 3(9 - a - b). Since the right-hand side is an integer, the left-hand side, i.e., b/2, must also be an integer. So we can brute-force and try b = 0, 2, 4, 6, 8 and see which one works. Trying one by one gives a = 10, b = 2 or a = 3, b = 6 as the only solutions with b in the range 0 <= b <= 9. Since a + b <= 9, we can reject the first solution. Doing so, we have $$\begin{cases}a = 3\\b = 6\end{cases}$$ as the only nonnegative integer solution to the equation.

Now, the constant term is $$\displaystyle \binom{9}{3,6,0} (2z)^3 \left(-\dfrac1{\sqrt z}\right)^6(z^3)^0 = \binom{9}3 2^3 = 672$$.

MaxWong May 8, 2022