7.2 Q-6

Exercise 7.2 — Question 6

This question applies the Binomial Theorem to find middle terms, specific terms, remainders, or last digits of large powers, depending on the exact sub-parts of Q6 in the FBISE textbook.

Key Concepts Required

General Term Formula

For the binomial expansion of $(a+b{)}^n$, the general term is:

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

Middle Term(s)

The total number of terms in $(a+b{)}^n$ is $n+1$.

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

Finding the Remainder Using the Binomial Theorem

To find the remainder when a large power is divided by a number, rewrite the base as a sum involving that divisor.

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:

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

$\therefore \text{Remainder}=1$

Finding the Last Digit Using the Binomial Theorem

To find the units digit of a large power, rewrite the base so that all but one term are divisible by $10$.

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 $10$, so:

$3^{100}=(-1{)}^{50}+10k=1+10k$

$\therefore \text{Last digit}=1$

Worked Strategy for Q6

1. Identify the type of sub-part: expansion, specific term, middle term, remainder, or last digit.
2. Write the general term $T_{r+1}=\left(\genfrac{}{}{0ex}{}{n}{r}\right)a^{n-r}{b}^r$.
3. For remainder/last digit problems, rewrite the base as $(\text{divisor}\pm 1{)}^n$ or $(10k\pm c{)}^n$.
4. Expand and isolate the constant term to read off the remainder or units digit.