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Binomial Theorem question

2025 · 3 Apr · Shift 1 · Q40

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Mathematics
Binomial Theorem
2025 · 3 Apr · Shift 1 · Q40

Binomial Theorem question

2025 · 3 Apr · Shift 1 · Q40

JEE MainMathematicsBinomial TheoremMCQ+4 / −1

The sum of all rational terms in the expansion of

(2+\sqrt{3})^8

is :

A
16923
B
18817
C
3763
D
33845

View written solutionFree

Correct answer: B

We need the sum of all rational terms in the expansion of

(2+\sqrt{3})^8.

General term in the binomial expansion is

T_{k+1}=\binom{8}{k}2^{8-k}(\sqrt{3})^k, \qquad k=0,1,2,\dots,8.

A term will be rational only when the power of $\sqrt{3}$ is even, i.e. when $k$ is even.

So rational terms correspond to

k=0,2,4,6,8.

Compute these terms:

For $k=0$ :

\binom{8}{0}2^8(\sqrt{3})^0=1\cdot 256=256.

For $k=2$ :

\binom{8}{2}2^6(\sqrt{3})^2=28\cdot 64\cdot 3=5376.

For $k=4$ :

\binom{8}{4}2^4(\sqrt{3})^4=70\cdot 16\cdot 9=10080.

For $k=6$ :

\binom{8}{6}2^2(\sqrt{3})^6=28\cdot 4\cdot 27=3024.

For $k=8$ :

\binom{8}{8}2^0(\sqrt{3})^8=1\cdot 81=81.

Add all rational terms:

256+5376+10080+3024+81=18817.

Therefore, the sum of all rational terms is

\boxed{18817}.

Comparing with the given options:

A: $16923$
B: $18817$
C: $3763$
D: $33845$

Hence the correct option is

\boxed{\text{B}}.

Comparison with stored correct answer: Stored correct answer is B, which matches our result.

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