A) 32
B) 33
C) 34
D) 35
Correct Answer: B
Solution :
If (r + 1)th term is the general term in the expansion of \[{{(x+\alpha )}^{n}}\]. \[{{T}_{r+1}}{{=}^{n}}{{C}_{r}}{{x}^{n-r}}{{\alpha }^{r}}\] (r+1th term of \[{{(\sqrt{3}+\sqrt[8]{5})}^{256}}\] i.e., \[{{T}_{r+1}}{{=}^{256}}{{C}_{r}}{{(3)}^{256-r)/2}}{{(5)}^{r/8}}\] Since, terms are integral, if \[\frac{256-r}{2}\] and \[\frac{r}{8}\] are both positive integers. i.e., r = 0,8, 16, 24, 32, ..., 256 Hence, total terms are 33. Alternate Solution \[{{(1+x)}^{27/5}}=1+\frac{27}{5}x+\frac{27}{5}\left( \frac{27}{5}-1 \right)\frac{{{x}^{2}}}{2!}\] \[+\frac{27}{5}\left( \frac{27}{5}-1 \right)\left( \frac{27}{5}-2 \right)\frac{{{x}^{3}}}{3!}\] \[+\frac{27}{5}\left( \frac{27}{5}-1 \right)\left( \frac{27}{5}-2 \right)\left( \frac{27}{5}-3 \right)\frac{{{x}^{4}}}{4!}\] \[+\frac{27}{5}\left( \frac{27}{5}-1 \right)\left( \frac{27}{5}-2 \right)\left( \frac{27}{5}-3 \right)\left( \frac{27}{5}-4 \right)\,\frac{{{x}^{5}}}{5!}\] \[+\frac{27}{5}\left( \frac{27}{5}-1 \right)\left( \frac{27}{5}-2 \right)\left( \frac{27}{5}-3 \right)\left( \frac{27}{5}-4 \right)\left( \frac{27}{5}-5 \right)\,\]\[\frac{{{x}^{6}}}{6!}\] \[+\frac{27}{5}\left( \frac{27}{5}-1 \right)\left( \frac{27}{5}-2 \right)\left( \frac{27}{5}-3 \right)\left( \frac{27}{5}-4 \right)\left( \frac{27}{5}-5 \right)\times \left( \frac{27}{5}-6 \right)\frac{{{x}^{7}}}{7!}+...\] Here, \[\frac{27}{5}-6\] is negative i.e., 8th term is negative In the expansion of \[{{(1+x)}^{27/5}}\].
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