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Use the first four terms of the expansion of (1 + 0.03)⁶ to estimate 1.03⁶…

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Use the first four terms of the expansion of $(1 + 0.03)^6$ to estimate $1.03^6$ . The exact value to six decimal places is $1.194052$ . How large is the error?

Need a reminder? The key ideaShow

For even $n$ the middle term is $T_{n/2+1}$ . For odd $n$ the middle terms are $T_{(n+1)/2}$ and $T_{(n+3)/2}$ . The term free of $x$ has the power $0$ of $x$ . For a small $x$ , $(1 + x)^n \approx 1 + nx + \dfrac{n(n - 1)}{2}x^2 + \dots$

The worked solution

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Use the first four terms of the expansion of (1 + 0.03)⁶ to estimate 1.03⁶…

The worked solution

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