Question

In the expansion of ${({2.2}^{\frac{2x}{3}}+2^{\frac{-x}{3}})}^n$ the sum of the last four binomial coefficients exceeds the sum of the first three binomial coefficients by 20 and if the middle term is-the numerically largest term, then x belongs to

• A. $(-2,\infty )$
• B. $(2,lo{g}_2\frac{1}{3})$
• C. $(-2,lo{g}_2\frac{1}{3})$
• D. $(-2,2lo{g}_2\frac{2}{3})$

Answer 1

The correct option is D $(-2,2lo{g}_2\frac{2}{3})$

Applying the above given condition we get
$({}^n{C}_n+{}^n{C}_{n-1}+{}^n{C}_{n-2}+{}^n{C}_{n-3})-({}^n{C}_0+{}^n{C}_1+{}^n{C}_2)=20$
By using properties of binomial coefficients the above expression is reduced to
${}^n{C}_{n-3}=20$
${}^n{C}_3=20$
${}^n{C}_3={}^6{C}_3$
Hence $n=6$.
It is given that the middle term is the largest term.
Hence $T_{3+1}$
$=T_4=$largest term.
Therefore $\frac{T_4}{T_3}>1$ which implies
${\frac{2}{3}}^2>2^x$
$2log2-2log3>xlog2$
$2-2lo{g}_{2}3>x$ ...(i)
Now $\frac{T_4}{T_5}>1$
Which is simplified as $2^x>\frac{1}{4}$
$xlog2>-2log2$
$x>-2$...(ii)
From i and ii we get
$xϵ(-2,2-2lo{g}_{2}3)$
Hence answer is Option D