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Binomial Theorem question

2003 · Shift 0 · Q106

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Mathematics
Binomial Theorem
2003 · Shift 0 · Q106

Binomial Theorem question

2003 · Shift 0 · Q106

JEE MainMathematicsBinomial TheoremMCQ+4 / −1

The number of integral terms in the expansion of

{\left( {\sqrt 3 + \sqrt[8]{5} } \right)^{256}}

is

A
35
B
32
C
33
D
34

View written solutionFree

Correct answer: C

General term of the binomial expansion

For

\left(\sqrt{3}+\sqrt[8]{5}\right)^{256},

the general term is

T_{r+1}=\binom{256}{r}(\sqrt{3})^{256-r}(\sqrt[8]{5})^r, \qquad r=0,1,2,\dots,256.

So,

T_{r+1}=\binom{256}{r}3^{\frac{256-r}{2}}\,5^{\frac{r}{8}}.

We want the term to be integral.

Condition for integrality

Since $\binom{256}{r}$ is always an integer, the term will be an integer only if both exponents

\frac{256-r}{2} \quad \text{and} \quad \frac{r}{8}

are integers.

That gives:

$\dfrac{r}{8}$ must be an integer $\Rightarrow r$ must be divisible by $8$ .
$\dfrac{256-r}{2}$ must be an integer. But $256$ is even, so this only requires $r$ to be even, which is automatically true if $r$ is divisible by $8$ .

Hence the only condition is:

r\equiv 0 \pmod{8}.

Count such values of $r$

We need multiples of $8$ from $0$ to $256$ inclusive:

r=0,8,16,\dots,256.

This is an arithmetic progression with number of terms

\frac{256-0}{8}+1=32+1=33.

Conclusion

Therefore, the number of integral terms is

\boxed{33}.

So the correct option is C.

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