Review question

# Can we use a binomial expansion to evaluate $\left(19\tfrac{3}{4}\right)^6$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7978

## Solution

Expand $(x-y)^6$ by the binomial theorem…

Using the binomial theorem, we have $(a+b)^n=\sum\limits_{k=0}^n \binom{n}{k}a^{n-k}b^k.$ Substituting in the values $n=6$, $a=x$ and $b=-y$, we get

\begin{align*} (x-y)^6 &= \binom{6}{0}x^6(-y)^0+\binom{6}{1}x^5(-y)^1+\binom{6}{2}x^4(-y)^2+\binom{6}{3}x^3(-y)^3 \\ &\quad + \binom{6}{4}x^2(-y)^4+\binom{6}{5}x^1(-y)^5+\binom{6}{6}x^0(-y)^6 \\ &= x^6(-y)^0 + \frac{6}{1} x^5(-y)^1 + \frac{6 \cdot 5}{2 \cdot 1} x^4(-y)^2 + \frac{6 \cdot 5 \cdot 4}{3 \cdot 2 \cdot 1} x^3(-y)^3 \\ &\quad + \frac{6 \cdot 5}{2 \cdot 1} x^2(-y)^4 + \frac{6}{1} x^1(-y)^5 + x^0(-y)^6 \\ &=x^6 -6x^5y+15x^4y^2-20x^3y^3+15x^2y^4-6xy^5+y^6 . \end{align*}

Alternatively we can use the fact that the 6th line of Pascal’s triangle is $1$, $6$, $15$, $20$, $15$, $6$, $1$ and just write the final result down, being careful with signs. (If you do this, it’s a good idea to state that you’ve used the $6^{\textrm{th}}$ line of Pascal’s triangle.)

…and use the result to evaluate $\left(19\tfrac{3}{4}\right)^6$ correct to the nearest thousand.

Using $x=20$ and $y=\frac{1}{4}$, we can substitute into the binomial expansion to get \begin{align*} \left(19\tfrac{3}{4}\right)^6 & = 20^6 - 6 \times 20^5 \times \frac{1}{4} + 15 \times 20^4 \times \left(\frac{1}{4}\right)^2 - 20 \times 20^3 \times \left(\frac{1}{4}\right)^3 \\ &\qquad \qquad +\, 15 \times 20^2 \times \left(\frac{1}{4}\right)^4 - 6 \times 20 \times \left(\frac{1}{4}\right)^5 + \left(\frac{1}{4}\right)^6. \end{align*}

We need this result to the nearest thousand, so we can save ourselves some work and ignore the very last term straight away.

We can then work backwards through the terms, calculating their rough size to see whether we can ignore them in the approximation.

It’s helpful to write numbers in their prime factorisation (to help with cancelling) or in powers of $10$ so we can keep tabs on how big the terms are.

• $\left(\dfrac{1}{4}\right)^6 = 2^{-12}$ — ignore

• $6\times 20\times \left(\dfrac{1}{4}\right)^5 = 3\times 10 \times 2^{-8}=\dfrac{30}{256}$ — ignore

• $15\times 20^2 \times \left(\dfrac{1}{4}\right)^4 =3\times 5 \times 10^2 \times 2^{-6}=\dfrac{1500}{64}$ — ignore

• $20^4\times \left(\dfrac{1}{4}\right)^3 = 10^4\times 2^{-2}=\dfrac{10000}{4}=2500$ — keep this.

The last three terms must be included in the approximation.

• $15\times 20^4\times \left(\dfrac{1}{4}\right)^2 = 3\times 5 \times 10^4=150\,000$

• $6\times 20^5 \times \dfrac{1}{4} = 2^4\times 3\times 10^5=4\,800\,000$

• $20^6 = 2^6 \times 10^6=64\,000\,000$.

So to the nearest thousand, \begin{align*} (19\tfrac{3}{4})^6 &\approx 64\,000\,000-4\,800\,000+150\,000-2500= 59\,347\,500 \\ &\approx 59\,348\,000. \end{align*}

The next term, $\dfrac{1500}{64}$, is positive, so we definitely need to round up.

A computer tells us that the exact answer is $\left(19\tfrac{3}{4}\right)^6=59\,347\,523.320556640625.$ It is interesting to note that including one more term would have given us the answer correct to the nearest whole number.