Let the coefficients of three consecutive terms T_r , T_r+1 and T_r+2 in the binomial expansion of (a + b)^12 be in a G.P. and let p be the number of all possible values of r . Let q be the sum of all rational terms in the binomial expansion of (\sqrt[4]3+\sqrt[3]4)^12 . Then p + q is equal to:

Binomial Theorem PYQ — JEE Main 2025 · 28 Jan · Shift 2 · Q32 with Solution

View written solutionFree

Correct answer: B

Find $p$ from the G.P. condition on coefficients in $(a+b)^{12}$

In the expansion of $(a+b)^{12}$ , the coefficient of the $(r+1)$ -th term is $\binom{12}{r}.$

So the three consecutive coefficients are $\binom{12}{r-1},\quad \binom{12}{r},\quad \binom{12}{r+1}$ if we index terms in the usual way, or equivalently for consecutive terms $T_r,T_{r+1},T_{r+2}$ the coefficients are consecutive binomial coefficients.
For three numbers to be in G.P., the square of the middle one must equal the product of the other two: $\binom{12}{r}^2=\binom{12}{r-1}\binom{12}{r+1}.$

Now use

\qquad \frac{\binom{12}{r+1}}{\binom{12}{r}}=\frac{12-r}{r+1}.$$ From $$\binom{12}{r}^2=\binom{12}{r-1}\binom{12}{r+1},$$ we get $$\frac{\binom{12}{r}}{\binom{12}{r-1}}=\frac{\binom{12}{r+1}}{\binom{12}{r}}.$$ Hence, $$\frac{13-r}{r}=\frac{12-r}{r+1}.$$ Cross-multiplying, $$ (13-r)(r+1)=r(12-r). $$ Expanding: $$13r+13-r^2-r=12r-r^2$$ $$12r+13-r^2=12r-r^2$$ $$13=0,$$ which is impossible. So **no** three consecutive binomial coefficients of $(a+b)^{12}$ are in G.P. Therefore, $$p=0.$$ --- 2. **Find $q$: sum of all rational terms in $(\sqrt[4]{3}+\sqrt[3]{4})^{12}$** Write $$x=\sqrt[4]{3}=3^{1/4}, \qquad y=\sqrt[3]{4}=4^{1/3}=2^{2/3}.$$ General term in the expansion is $$T_{k+1}=\binom{12}{k}x^{12-k}y^k.

Since $4^{k/3}=2^{2k/3}$ , rationality depends on both exponents being integers.

For the term to be rational, we need:

$\dfrac{12-k}{4}$ to be an integer,
$\dfrac{k}{3}$ to be an integer.

Thus, $12-k\equiv 0 \pmod 4 \Rightarrow k\equiv 0 \pmod 4,$ and $k\equiv 0 \pmod 3.$ So $k$ must be a multiple of $\operatorname{lcm}(3,4)=12$ .

Since $0\le k\le 12$ , possible values are $k=0,\ 12.$

Now compute these terms:

For $k=0$ : $T_1=\binom{12}{0}(\sqrt[4]{3})^{12}=3^3=27.$
For $k=12$ : $T_{13}=\binom{12}{12}(\sqrt[3]{4})^{12}=4^4=256.$

Hence, $q=27+256=283.$

Compute $p+q$

$p+q=0+283=283.$

So the correct option is $\boxed{\text{B: }283}.$

Comparison with stored answer

Stored correct answer: B
Derived answer: B

They agree.

More from Binomial Theorem

View written solutionFree

Correct answer: B

Find $p$ from the G.P. condition on coefficients in $(a+b)^{12}$

In the expansion of $(a+b)^{12}$ , the coefficient of the $(r+1)$ -th term is $\binom{12}{r}.$

Now use

\qquad \frac{\binom{12}{r+1}}{\binom{12}{r}}=\frac{12-r}{r+1}.$$ From $$\binom{12}{r}^2=\binom{12}{r-1}\binom{12}{r+1},$$ we get $$\frac{\binom{12}{r}}{\binom{12}{r-1}}=\frac{\binom{12}{r+1}}{\binom{12}{r}}.$$ Hence, $$\frac{13-r}{r}=\frac{12-r}{r+1}.$$ Cross-multiplying, $$ (13-r)(r+1)=r(12-r). $$ Expanding: $$13r+13-r^2-r=12r-r^2$$ $$12r+13-r^2=12r-r^2$$ $$13=0,$$ which is impossible. So **no** three consecutive binomial coefficients of $(a+b)^{12}$ are in G.P. Therefore, $$p=0.$$ --- 2. **Find $q$: sum of all rational terms in $(\sqrt[4]{3}+\sqrt[3]{4})^{12}$** Write $$x=\sqrt[4]{3}=3^{1/4}, \qquad y=\sqrt[3]{4}=4^{1/3}=2^{2/3}.$$ General term in the expansion is $$T_{k+1}=\binom{12}{k}x^{12-k}y^k.

Since $4^{k/3}=2^{2k/3}$ , rationality depends on both exponents being integers.

For the term to be rational, we need:

$\dfrac{12-k}{4}$ to be an integer,
$\dfrac{k}{3}$ to be an integer.

Thus, $12-k\equiv 0 \pmod 4 \Rightarrow k\equiv 0 \pmod 4,$ and $k\equiv 0 \pmod 3.$ So $k$ must be a multiple of $\operatorname{lcm}(3,4)=12$ .

Since $0\le k\le 12$ , possible values are $k=0,\ 12.$

Now compute these terms:

For $k=0$ : $T_1=\binom{12}{0}(\sqrt[4]{3})^{12}=3^3=27.$
For $k=12$ : $T_{13}=\binom{12}{12}(\sqrt[3]{4})^{12}=4^4=256.$

Hence, $q=27+256=283.$

Compute $p+q$

$p+q=0+283=283.$

So the correct option is $\boxed{\text{B: }283}.$

Comparison with stored answer

Stored correct answer: B
Derived answer: B

They agree.

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