In the binomial expansion of ${(^{3} \sqrt{2} + \frac{1}{^{3} \sqrt{3}})}^{n}$ , the ratio of the $7 t h$ terms from the beginning to the $7 t h$ terms from the end is $1 : 6$ , find $n$ .

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Solution

${(^{3} \sqrt{2} + \frac{1}{^{3} \sqrt{3}})}^{n}$

$T_{7} =^{n} C_{6} (2^{1 / 3})^{n - 6} (3^{- 1 / 3})^{6}$

$T_{from end} =^{n} C_{n - 8} (2^{1 / 3})^{6} (3^{- 1 / 3})^{n - 6}$

$\frac{1}{6} = \frac{^{n} C_{6} 2^{\frac{n - 6}{3}} \times 3^{\frac{- 6}{3}}}{^{n} C_{n - 6} 2^{2} 3^{- ⎛ ⎝ \frac{n - 6}{3} ⎞ ⎠}}$

$\frac{1}{6} = 2^{\frac{n - 6}{3} - \frac{6}{3}} \times 3^{\frac{n - 6}{3} - \frac{6}{3}}$

$\frac{1}{6} = (6)^{\frac{n - 6}{3} - \frac{6}{3}}$

$6^{- 1} = (6)^{\frac{n - 6}{3} - \frac{6}{3}}$

$- 1 = \frac{n - 6}{3} - \frac{6}{3}$

$- 1 + 2 = \frac{n - 6}{3}$

$1 \times 3 = n - 6$
$9 = n$

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In the binomial expansion of (3√2+13√3)n, the ratio of the 7th terms from the beginning to the 7th terms from the end is 1:6, find n.

(3√2+13√3)nT7=nC6(21/3)n−6(3−1/3)6 Tfrom end=nCn−8(21/3)6(3−1/3)n−616=nC62n−63×3−63nCn−6 22 3−⎛⎝n−63⎞⎠16=2n−63−63×3n−63−63 16=(6)n−63−63 6−1=(6)n−63−63 −1=n−63−63−1+2=n−63 1×3=n−6 9=n

In the binomial expansion of ${(^{3} \sqrt{2} + \frac{1}{^{3} \sqrt{3}})}^{n}$ , the ratio of the $7 t h$ terms from the beginning to the $7 t h$ terms from the end is $1 : 6$ , find $n$ .