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Cobedangiu   4
N Apr 5, 2025 by Lankou
Find the integer coefficients after expanding Newton's binomial:
$$(\frac{3}{2}-\frac{2}{3}x^2)^n (n \in Z)$$
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Cobedangiu
Apr 4, 2025
Lankou
Apr 5, 2025
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Cobedangiu
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Cobedangiu
#1 h  Apr 4, 2025, 6:20 AM
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Find the integer coefficients after expanding Newton's binomial:
$$(\frac{3}{2}-\frac{2}{3}x^2)^n (n \in Z)$$
This post has been edited 1 time. Last edited by Cobedangiu, Apr 4, 2025, 6:20 AM
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Cobedangiu
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#2 h  Apr 4, 2025, 8:36 AM
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Lankou
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Lankou
#3 h  Apr 4, 2025, 12:06 PM • 1 Y
Y by Cobedangiu
$(\frac{3}{2}-\frac{2}{3}x^2)^n =\sum_{k=0}^n {n\choose k} \cdot \left(\frac{3}{2}\right)^k \left(-\frac{2x^2}{3}\right)^{n-k}$
The coefficient is an integer when $n-k=k$
Coefficient$= {n\choose \frac{n}{2}}$
This post has been edited 1 time. Last edited by Lankou, Apr 4, 2025, 12:11 PM
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Cobedangiu
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Cobedangiu
#4 h  Apr 5, 2025, 9:45 AM
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Lankou wrote:
$(\frac{3}{2}-\frac{2}{3}x^2)^n =\sum_{k=0}^n {n\choose k} \cdot \left(\frac{3}{2}\right)^k \left(-\frac{2x^2}{3}\right)^{n-k}$
The coefficient is an integer when $n-k=k$
Coefficient$= {n\choose \frac{n}{2}}$

integer coefficients? ${n\choose \frac{n}{2}}$ not integer
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Lankou
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#5 h  Apr 5, 2025, 11:40 AM • 1 Y
Y by Cobedangiu
${n\choose \frac{n}{2}}$ always an integer
By the way it should be $(-1)^{\frac{n}{2}}{n\choose \frac{n}{2}}$
This post has been edited 3 times. Last edited by Lankou, Apr 5, 2025, 1:29 PM
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