The number of integral terms in the expansion of (√3+[8]5)^256 is:

Question

The number of integral terms in the expansion of $\left(\sqrt3+\sqrt[8]5\right)^{256}$ is:

Accepted Answer

General term = $${}^{256}{C_r}.{\left( {\sqrt 3 } \right)^{256 - r}}.{\left( {\root 8 \of 5 } \right)^r}$$

= $${}^{256}{C_r}.{\left( 3 \right)^{{{256 - r} \over 2}}}.{\left( 5 \right)^{{r \over 8}}}$$

When $${{{256 - r} \over 2}}$$ is integer then $${\left( 3 \right)^{{{256 - r} \over 2}}}$$ is integer.

And when $${{r \over 8}}$$ is integer then $${\left( 5 \right)^{{r \over 8}}}$$ is integer.

Entire general term will be integer when $${{{256 - r} \over 2}}$$ and $${{r \over 8}}$$ both are integer.

$${{{256 - r} \over 2}}$$ is integer when r = 0, 2, 4, 6, ......, 256

$${{r \over 8}}$$ is integer when r = 0, 8, 16 ,......., 256

Now both $${{{256 - r} \over 2}}$$ and $${{r \over 8}}$$ will be integer when r = 0, 8, 16, ...., 256 (This is an AP)

$$\therefore$$ 256 = 0 + (n - 1)8 using formula of AP, tn = a + (n - 1)d

$$\therefore$$ n = $${{256} \over 8} + 1$$ = 32 + 1 = 33

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