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Binomial Theorem question

2025 · 23 Jan · Shift 1 · Q46

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Binomial Theorem
2025 · 23 Jan · Shift 1 · Q46

Binomial Theorem question

2025 · 23 Jan · Shift 1 · Q46

JEE MainMathematicsBinomial TheoremNumerical+4 / −1

The sum of all rational terms in the expansion of

\left(1+2^{1 / 3}+3^{1 / 2}\right)^6

is equal to

\underline{\hspace{2cm}}

.

Numerical answer

View written solutionFree

Correct answer: 612

We need the sum of all rational terms in the expansion of $\left(1+2^{1/3}+3^{1/2}\right)^6.$
Write the general term using multinomial expansion: $T=\frac{6!}{a!b!c!}(1)^a\left(2^{1/3}\right)^b\left(3^{1/2}\right)^c,$ where $a+b+c=6,$ and $a,b,c\ge 0$ .

So, $T=\frac{6!}{a!b!c!}2^{b/3}3^{c/2}.$

For this term to be rational, both exponents must make rational powers:

$2^{b/3}$ is rational only if $\frac b3$ is an integer, i.e. $b\equiv 0\pmod 3$ .
$3^{c/2}$ is rational only if $\frac c2$ is an integer, i.e. $c\equiv 0\pmod 2$ .

Thus we need: $b\in\{0,3,6\},\qquad c\in\{0,2,4,6\},$ with $a=6-b-c\ge 0$ .

Now list all valid triples $(a,b,c)$ .

Case 1: $b=0$

Then $c=0,2,4,6$ are possible.

$(a,b,c)=(6,0,0)$ $T=\frac{6!}{6!0!0!}=1$
$(4,0,2)$ $T=\frac{6!}{4!0!2!}3^{1}=\frac{720}{24\cdot 2}\cdot 3=15\cdot 3=45$
$(2,0,4)$ $T=\frac{6!}{2!0!4!}3^{2}=\frac{720}{2\cdot 24}\cdot 9=15\cdot 9=135$
$(0,0,6)$ $T=\frac{6!}{0!0!6!}3^{3}=1\cdot 27=27$

Sum from this case: $1+45+135+27=208$

Case 2: $b=3$

Then $c=0,2$ are possible.

$(3,3,0)$ $T=\frac{6!}{3!3!0!}2^{1}=\frac{720}{6\cdot 6}\cdot 2=20\cdot 2=40$
$(1,3,2)$ $T=\frac{6!}{1!3!2!}2^{1}3^{1}=\frac{720}{1\cdot 6\cdot 2}\cdot 6=60\cdot 6=360$

Sum from this case: $40+360=400$

Case 3: $b=6$

Then $c=0$ only.

$(0,6,0)$ $T=\frac{6!}{0!6!0!}2^{2}=1\cdot 4=4$

Sum from this case: $4$

Therefore total sum of all rational terms is $208+400+4=612.$
Comparing with the stored correct answer $612$ , they match.

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