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Find n in the binomial (2 3+1/(3 3))^n ,...

Find `n` in the binomial `(2 3+1/(3 3))^n` , if the ration 7th term from the beginning to the 7 term from the end `1/6dot`

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The correct Answer is:
`n = 9`
`1/6 = (.^(n)C_(6)(2^(1//3))^(n-6)(3^(-1//3))^(6))/(.^(n)C_(n-6)(2^(1//3))^(n-6)(3^(-1//3))^(n-6))`
`6^(-1) = 6^(-4) xx 6^(n//3)= 6^(n//3-4) rArr n/3 - 4 = - 1 rArr n = 9`
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CENGAGE-BINOMIAL THEOREM-Exercise 8.1
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  6. Prove that sum(r=0)^(n) (-1)^(r)""^(n)C(r)[1/(2^(r))+(3^(r))/(2^(2r...

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  7. Find n in the binomial (2 3+1/(3 3))^n , if the ration 7th term from t...

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  10. Represent cos 6 theta in terms of cos theta.

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  12. Find the value of (sqrt(2)+1)^6-(sqrt(2)-1)^6dot

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  16. Find the middle term in the expansion of (x^2+1/(x^2)+2^n)dot

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