Can we find an approximation to $(1.0099)^6$?

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Obtain the first \(4\) terms in the expansion of \((1 + p)^6\) in ascending powers of \(p\).

By substituting \(p = x - x^2\) obtain the expansion of \((1 + x - x^2)^6\) as far as the term in \(x^2\).

Find a value for \(x\) which would enable you to calculate \((1.0099)^6\).