further work with coefficients binomial (1 Viewer)

Halo189 · Apr 18, 2015

The coefficients of x^4 and x^5 in the expansion of (3 − x)^n are equal in magnitude but opposite in sign. Find the value of n

answer: n=19

Thanks

InteGrand · Apr 18, 2015

Halo189 said:
The coefficients of x^4 and x^5 in the expansion of (3 − x)^n are equal in magnitude but opposite in sign. Find the value of n

answer: n=19

Thanks

By the binomial theorem, $(3-x)^n=\sum_{k-0}^{n}\binom{n}{k}3^{n-k}(-x)^k=\sum_{k-0}^{n}\binom{n}{k}3^{n-k}(-1)^kx^k$ .

The (-1)^k tells us that the signs of the terms alternate, so we only need to focus on the magnitude. The above formula tells us that the coefficient of x^k has magnitude $|T_k|=\binom{n}{k}3^{n-k}=\frac{n!}{k!(n-k)!}\cdot 3^{n-k}$ .

Therefore, the absolute value of the coefficient of x⁴ and x⁵ are $|T_4|=\frac{n!}{4!(n-4)!}\cdot 3^{n-4}$ and $|T_5|=\frac{n!}{5!(n-5)!}\cdot 3^{n-5}$ .

Since we want these to be equal, we have

$\frac{n!}{4!(n-4)!}\cdot 3^{n-4} =\frac{n!}{5!(n-5)!}\cdot 3^{n-5}$

$\Rightarrow \frac{\frac{n!}{4!(n-4)!}}{\frac{n!}{5!(n-5)!}}=\frac{3^{n-5}}{3^{n-4}}$

$\Rightarrow \frac{5!(n-5)!}{4!(n-4)!}=\frac{3^{n-5}}{3^{n-4}}$

$\Rightarrow \frac{5}{n-4}=\frac{1}{3}$

$\Rightarrow \frac{n-4}{5}=3$

$\Rightarrow n=5\times 3 + 4 = 19.$

Halo189 · Apr 19, 2015

InteGrand said:
By the binomial theorem, $(3-x)^n=\sum_{k-0}^{n}\binom{n}{k}3^{n-k}(-x)^k=\sum_{k-0}^{n}\binom{n}{k}3^{n-k}(-1)^kx^k$ .

The (-1)^k tells us that the signs of the terms alternate, so we only need to focus on the magnitude. The above formula tells us that the coefficient of x^k has magnitude $|T_k|=\binom{n}{k}3^{n-k}=\frac{n!}{k!(n-k)!}\cdot 3^{n-k}$ .

Therefore, the absolute value of the coefficient of x⁴ and x⁵ are $|T_4|=\frac{n!}{4!(n-4)!}\cdot 3^{n-4}$ and $|T_5|=\frac{n!}{5!(n-5)!}\cdot 3^{n-5}$ .

Since we want these to be equal, we have

$\frac{n!}{4!(n-4)!}\cdot 3^{n-4} =\frac{n!}{5!(n-5)!}\cdot 3^{n-5}$

$\Rightarrow \frac{\frac{n!}{4!(n-4)!}}{\frac{n!}{5!(n-5)!}}=\frac{3^{n-5}}{3^{n-4}}$

$\Rightarrow \frac{5!(n-5)!}{4!(n-4)!}=\frac{3^{n-5}}{3^{n-4}}$

$\Rightarrow \frac{5}{n-4}=\frac{1}{3}$

$\Rightarrow \frac{n-4}{5}=3$

$\Rightarrow n=5\times 3 + 4 = 19.$

Thank you

further work with coefficients binomial (1 Viewer)

Halo189

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InteGrand

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