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

The

binomial theorem says that, if

\(x\) is a real number and

\(n\) is a positive integer, then

\[
(1+x)^n = \sum_{i=0}^n \binom{n}{i} x^i
\] where

\[\begin{align*}
\binom{n}{0} &= 1, \\
\binom{n}{i} &= \frac{n(n-1) \dotsm (n-(i-1))}{i(i-1) \dotsm 1} \quad\text{for each integer $1 \le i \le n$.}
\end{align*}\]
By applying this to expression

\((1+p)^6\) and ignoring terms beyond the fourth, we see that

\[\begin{align*}
(1+p)^6 &= \binom{6}{0} p^0 + \binom{6}{1} p^1 + \binom{6}{2} p^2 + \binom{6}{3} p^3 + \dotsb \\
&= 1 + \frac{6}{1} p + \frac{6 \times 5}{2 \times 1} p^2 + \frac{6 \times 5 \times 4}{3 \times 2 \times 1} p^3 + \dotsb \\
&= 1 + 6p + 15p^2 + 20p^3 + \dotsb.
\end{align*}\]
By substituting \(p = x - x^2\) obtain the expansion of \((1 + x - x^2)^6\) as far as the term in \(x^2\).

By taking

\(p = x - x^2\) in the above expansion, we have that

\[
(1+x-x^2)^6 = 1 + 6(x-x^2) + 15(x-x^2)^2 + 20(x-x^2)^3 + \dotsb.
\] Ignoring the terms corresponding to

\(p^3\),

\(p^4\),… we have, up to and including the term featuring

\(x^2\),

\[\begin{align*}
(1+x-x^2)^6 &= 1 + 6(x-x^2) + 15x^2 + \dotsb \\
&= 1 + 6x - 6x^2 + 15x^2 + \dotsb \\
&= 1 + 6x + 9x^2 + \dotsb.
\end{align*}\]
Find a value for \(x\) which would enable you to calculate \((1.0099)^6\).

If we intend to use a binomial expansion of \((1+x-x^2)^6\), it would be sensible to find some \(x\) such that \[
1+x-x^2 = 1.0099.
\] That is, \(x\) must be such that \[
x - x^2 = 0.0099.
\] However, as \[
0.0099 = 0.01 - 0.0001 = 0.01 - (0.01)^2,
\] we can simply take \(x = 0.01\).

Alternatively, we could solve the quadratic equation \(x^2 - x + 0.0099 = 0\).

By applying this solution \(x=0.01\) to the approximation from the previous part \[
(1+x-x^2)^6 \approx 1 + 6x + 9x^2
\] which improves as \(x\) tends to zero, we get \[
(1.0099)^6 \approx 1 + 6 \times 0.01 + 9 \times (0.01)^2 = 1 + 0.06 + 0.0009 = 1.0609.
\] In fact, \[
(1.0099)^6 = 1.0608897,
\] and so the error in the approximation is tiny: it is less than \(0.001\)%.