A ratio of the 5th term from the beginning to the 5th term from the end in the binomial expansion of ( 2^1/3 + 1 \over 2( 3 )^1/3 )^10 is :

Binomial Theorem PYQ — JEE Main 2019 · 12 Jan · Shift 1 · Q24 with Solution

View written solutionFree

Correct answer: D

Identify the general term

For the expansion of

(a+b)^{10},

the $(r+1)$ -th term from the beginning is

T_{r+1}=\binom{10}{r}a^{10-r}b^r.

Here,

a=2^{1/3}, \qquad b=\frac{1}{2\,3^{1/3}}.

Find the 5th term from the beginning

The 5th term corresponds to $r=4$ :

T_5=\binom{10}{4}a^6b^4.

So,

T_5=\binom{10}{4}(2^{1/3})^6\left(\frac{1}{2\,3^{1/3}}\right)^4.

Now simplify:

(2^{1/3})^6=2^2=4,

and

\left(\frac{1}{2\,3^{1/3}}\right)^4=\frac{1}{2^4\,3^{4/3}}=\frac{1}{16\,3^{4/3}}.

Thus,

T_5=\binom{10}{4}\cdot \frac{4}{16\,3^{4/3}}=\binom{10}{4}\cdot \frac{1}{4\,3^{4/3}}.

Find the 5th term from the end

In $(a+b)^{10}$ , total number of terms is $11$ .

So, the 5th term from the end is the $(11-5+1)=7$ th term from the beginning. That means $r=6$ .

Hence,

T_7=\binom{10}{6}a^4b^6.

So,

T_7=\binom{10}{6}(2^{1/3})^4\left(\frac{1}{2\,3^{1/3}}\right)^6.

Now simplify:

(2^{1/3})^4=2^{4/3},

and

\left(\frac{1}{2\,3^{1/3}}\right)^6=\frac{1}{2^6\,3^2}= rac{1}{64\cdot 9}= rac{1}{576}.

Thus,

T_7=\binom{10}{6}\cdot \frac{2^{4/3}}{64\,3^2}.

Take the ratio $T_5:T_7$

Since

\binom{10}{4}=\binom{10}{6},

these cancel.

Therefore,

\frac{T_5}{T_7}= rac{a^6b^4}{a^4b^6}=\frac{a^2}{b^2}=\left(\frac{a}{b}\right)^2.

Now,

\frac{a}{b}=\frac{2^{1/3}}{\frac{1}{2\,3^{1/3}}}=2^{1/3}\cdot 2\,3^{1/3}=2^{4/3}3^{1/3}.

Hence,

\left(\frac{a}{b}\right)^2=2^{8/3}3^{2/3}.

Rewrite:

2^{8/3}3^{2/3}=4\cdot 2^{2/3}3^{2/3}=4(6)^{2/3}=4\sqrt[3]{36}.

So,

T_5:T_7=4\sqrt[3]{36}:1.

Match with the options

This is exactly:

\boxed{4(36)^{1/3}:1}

which is Option D.

Compare with stored correct answer

Stored correct answer: D

Our derived answer: D

So they agree.

More from Binomial Theorem

View written solutionFree

Correct answer: D

Identify the general term

For the expansion of

(a+b)^{10},

the $(r+1)$ -th term from the beginning is

T_{r+1}=\binom{10}{r}a^{10-r}b^r.

Here,

a=2^{1/3}, \qquad b=\frac{1}{2\,3^{1/3}}.

Find the 5th term from the beginning

The 5th term corresponds to $r=4$ :

T_5=\binom{10}{4}a^6b^4.

So,

T_5=\binom{10}{4}(2^{1/3})^6\left(\frac{1}{2\,3^{1/3}}\right)^4.

Now simplify:

(2^{1/3})^6=2^2=4,

and

\left(\frac{1}{2\,3^{1/3}}\right)^4=\frac{1}{2^4\,3^{4/3}}=\frac{1}{16\,3^{4/3}}.

Thus,

T_5=\binom{10}{4}\cdot \frac{4}{16\,3^{4/3}}=\binom{10}{4}\cdot \frac{1}{4\,3^{4/3}}.

Find the 5th term from the end

In $(a+b)^{10}$ , total number of terms is $11$ .

So, the 5th term from the end is the $(11-5+1)=7$ th term from the beginning. That means $r=6$ .

Hence,

T_7=\binom{10}{6}a^4b^6.

So,

T_7=\binom{10}{6}(2^{1/3})^4\left(\frac{1}{2\,3^{1/3}}\right)^6.

Now simplify:

(2^{1/3})^4=2^{4/3},

and

\left(\frac{1}{2\,3^{1/3}}\right)^6=\frac{1}{2^6\,3^2}= rac{1}{64\cdot 9}= rac{1}{576}.

Thus,

T_7=\binom{10}{6}\cdot \frac{2^{4/3}}{64\,3^2}.

Take the ratio $T_5:T_7$

Since

\binom{10}{4}=\binom{10}{6},

these cancel.

Therefore,

\frac{T_5}{T_7}= rac{a^6b^4}{a^4b^6}=\frac{a^2}{b^2}=\left(\frac{a}{b}\right)^2.

Now,

\frac{a}{b}=\frac{2^{1/3}}{\frac{1}{2\,3^{1/3}}}=2^{1/3}\cdot 2\,3^{1/3}=2^{4/3}3^{1/3}.

Hence,

\left(\frac{a}{b}\right)^2=2^{8/3}3^{2/3}.

Rewrite:

2^{8/3}3^{2/3}=4\cdot 2^{2/3}3^{2/3}=4(6)^{2/3}=4\sqrt[3]{36}.

So,

T_5:T_7=4\sqrt[3]{36}:1.

Match with the options

This is exactly:

\boxed{4(36)^{1/3}:1}

which is Option D.

Compare with stored correct answer

Stored correct answer: D

Our derived answer: D

So they agree.

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