Binomial Theorem Question - None

Question

The ratio of the 5th term from the beginning to the 5th term from the end in the binomial expansion of $\left(2^{\frac{1}{3}}+\frac{1}{2(3)^{\frac{1}{3}}}\right)^{10}$is

Accepted Answer

Since, rth term from the end in the expansion of a binomial (x + a)ⁿ is same as the (n - r + 2)th term from the beginning in the expansion of same binomial.
$ herefore$ Required ratio = $\frac{T_{5}}{T_{10-5+2}}=\frac{T_{5}}{T_{7}}=\frac{T_{4+1}}{T_{6+1}}$
$\Rightarrow \frac{T_{5}}{T_{10-5+2}}=\frac{^{10} C_{4}\left(2^{1 / 3} ight)^{10-4}\left(\frac{1}{2(3)^{1 / 3}} ight)^{4}}{^{10} C_{6}\left(2^{1 / 3} ight)^{10-6}\left(\frac{1}{2(3)^{1 / 3}} ight)^{6}}$ [$\because$ T_r+1 = ⁿC_rx^n-ra^r]
$=\frac{2^{6 / 3}\left(2(3)^{1 / 3} ight)^{6}}{2^{4 / 3}\left(2(3)^{1 / 3} ight)^{4}}$ [$\because$ ¹⁰C₄ = ¹⁰C₆]
= 2^{6/3 - 4/3} (2(3)^1/3)^6-4
= $2^{2/{3}} \cdot 2^{2} \cdot 3^{2/{3}}=4(6)^{2/{3}}=4(36)^{1 / 3}$
So, the required ratio is 4(36)^1/3 : 1

Question:

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