7.2 Q-10

Exercise 7.2 — Question 10

This question applies the Binomial Theorem to practical problems such as finding approximate values, remainders, and last digits of large powers.

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Key Formula Reminder

The general term of $(a+b{)}^n$ is:

$T_{r+1}=\left(\genfrac{}{}{0ex}{}{n}{r}\right)a^{n-r}{b}^r,r=0,1,2,\dots ,n$

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Technique 1 — Approximate Values

Write the expression in the form $(1+x{)}^n$ where $|x|$ is small, then expand and keep terms up to the required accuracy.

Example: Find an approximate value of $(1.02{)}^{10}$.

$(1.02{)}^{10}=(1+0.02{)}^{10}=\left(\genfrac{}{}{0ex}{}{10}{0}\right)(0.02{)}^0+(\genfrac{}{}{0ex}{}{10}{1})(0.02{)}^1+\left(\genfrac{}{}{0ex}{}{10}{2}\right)(0.02{)}^2+\cdots$

$\approx 1+10(0.02)+45(0.0004)+\cdots =1+0.2+0.018+\cdots \approx 1.218$

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Technique 2 — Finding Remainders

Express the base as $(\text{multiple of divisor}\pm 1{)}^n$, expand, and identify the remainder.

Example: Find the remainder when $7^{100}$ is divided by $6$.

$7^{100}=(6+1{)}^{100}={\sum }_{r=0}^{100}(\genfrac{}{}{0ex}{}{100}{r})6^r\cdot 1^{100-r}$

Every term with $r\ge 1$ contains a factor of $6$, so:

$7^{100}=6k+\left(\genfrac{}{}{0ex}{}{100}{0}\right)\cdot 1=6k+1$

Remainder $=1$.

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Technique 3 — Last Digit (Units Digit)

Express the base so that the last digit pattern repeats, then use the Binomial Theorem.

Example: Find the last digit of $3^{100}$.

$3^{100}=(3^2{)}^{50}=9^{50}=(10-1{)}^{50}$

$={\sum }_{r=0}^{50}\left(\genfrac{}{}{0ex}{}{50}{r}\right)10^r(-1{)}^{50-r}$

All terms with $r\ge 1$ are divisible by $10$ (contribute $0$ to the last digit). The $r=0$ term is:

$\left(\genfrac{}{}{0ex}{}{50}{0}\right)\cdot 10^0\cdot (-1{)}^{50}=1$

Last digit $=1$.

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Middle Terms

• If $n$ is even: one middle term $T_{\frac{n}{2}+1}$
• If $n$ is odd: two middle terms $T_{\frac{n+1}{2}}$ and $T_{\frac{n+3}{2}}$

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