To solve $(1.99)^5$ using the Binomial Theorem, follow these steps:

Step 1: Recognize the Binomial Expansion Formula

The binomial theorem states: $(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$

Here:

• $a = 2$ (the nearest integer to $1.99$),
• $b = -0.01$ (since $1.99 = 2 - 0.01$),
• $n = 5$.

Thus: $(1.99)^5 = (2 - 0.01)^5$

Step 2: Expand the Expression

Using the binomial theorem: $(2 - 0.01)^5 = \sum_{k=0}^5 \binom{5}{k} 2^{5-k} (-0.01)^k$

Write each term:

1. For $k = 0$: $\binom{5}{0} 2^5 (-0.01)^0 = 1 \cdot 32 \cdot 1 = 32$
2. For $k = 1$: $\binom{5}{1} 2^4 (-0.01)^1 = 5 \cdot 16 \cdot (-0.01) = -0.8$
3. For $k = 2$: $\binom{5}{2} 2^3 (-0.01)^2 = 10 \cdot 8 \cdot 0.0001 = 0.008$
4. For $k = 3$: $\binom{5}{3} 2^2 (-0.01)^3 = 10 \cdot 4 \cdot (-0.000001) = -0.00004$
5. For $k = 4$: $\binom{5}{4} 2^1 (-0.01)^4 = 5 \cdot 2 \cdot 0.00000001 = 0.0000001$
6. For $k = 5$: $\binom{5}{5} 2^0 (-0.01)^5 = 1 \cdot 1 \cdot (-0.0000000001) = -0.0000000001$

Step 3: Approximate by Summing the Terms

Now, add up all the significant terms (stopping when terms become negligible):

$(1.99)^5 \approx 32 - 0.8 + 0.008 - 0.00004$

Step 4: Calculate the Result

Performing the addition: $(1.99)^5 \approx 31.208 - 0.00004 \approx 31.20796$

Final Answer:

$(1.99)^5 \approx 31.208$

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Would you like further clarifications or a deeper breakdown of the process?

Related Questions:

1. How can the binomial theorem be applied to negative or fractional powers?
2. What are common approximation methods for powers close to integers?
3. How accurate is this method compared to a calculator’s exact result?
4. Can this expansion be used for estimating roots instead of powers?
5. What is the importance of truncating terms in practical calculations?

Tip:

When using the binomial theorem for approximations, consider truncating the series after terms that are smaller than your required precision.