Discussion
Using two terms of its Maclaurin series, \( \sqrt{1.02} \) is approximately
Answer: B
The binomial expansion for small \( t \) is \( (1+t)^{1/2} \approx 1 + t/2 \). With \( t = 0.02 \),
\[ \sqrt{1.02} \;\approx\; 1 + \frac{0.02}{2} \;=\; 1.01. \]
The next term is \( -t^{2}/8 = -0.00005 \), so the true value is 1.00995 and the two-term estimate is high by five parts in a million
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