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The number of integral terms ${(1 + \sqrt{2})}^{19}$ are

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answer is B.

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Detailed Solution

${(1 + \sqrt{2})}^{19} = \sum_{r = 0}^{19} (^{19} C_{r} 2^{\frac{r}{2}}) for integral terms substitute r = 0, 2, 4, 6,, 8, 10, 12, 14, 16, 18, Number of integral terms = 10$

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