A) 128
B) 129
C) 130
D) 131
Correct Answer: B
Solution :
Here\[\left[ \because \,\frac{^{n}{{C}_{k}}}{^{n}{{C}_{k-1}}}=\frac{n-k+1}{k} \right]\], a power of 2, whereas the power of 7 is \[\frac{1}{8}={{2}^{-3}}\] Now first term \[^{1024}{{C}_{0}}{{\left( {{5}^{1/2}} \right)}^{1024}}={{5}^{512}}\](integer) And after 8 terms, the 9th term \[{{=}^{\,\,\,1024}}{{C}_{8}}{{({{5}^{1/2}})}^{1016}}{{({{7}^{1/8}})}^{8}}\] = an integer Again, 17th term =\[^{1024}{{C}_{16}}{{({{5}^{1/2}})}^{1008}}{{({{7}^{1/8}})}^{16}}\] = An integer. Continuing like this, we get an A.P. 1, 9, 17, ...., 1025, because 1025th term = the last term in the expansion \[={{\,}^{1024}}{{C}_{1024}}{{\left( {{7}^{1/8}} \right)}^{1024}}={{7}^{128}}\](an integer) If n is the number of terms of above A.P. we have \[1025={{T}_{n}}=1+(n-1)8\,\,\,\Rightarrow n=129\].
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