Q.Using binomial theorem, evaluate $(102)^5$.

Concept: Binomial Theorem Expansion — rewrite $102$ as $100 + 2$ to use the expansion $(a+b)^n$.

Step 1: Write $(102)^5 = (100 + 2)^5$.

Step 2: Apply the binomial theorem:

Step 3: Compute each term:

• $k=0$: $\binom{5}{0}100^5 = 10^{10} = 10\,000\,000\,000$
• $k=1$: $\binom{5}{1}100^4 \cdot 2 = 5 \cdot 10^8 \cdot 2 = 1\,000\,000\,000$
• $k=2$: $\binom{5}{2}100^3 \cdot 4 = 10 \cdot 10^6 \cdot 4 = 40\,000\,000$
• $k=3$: $\binom{5}{3}100^2 \cdot 8 = 10 \cdot 10^4 \cdot 8 = 800\,000$
• $k=4$: $\binom{5}{4}100^1 \cdot 16 = 5 \cdot 100 \cdot 16 = 8\,000$
• $k=5$: $\binom{5}{5}100^0 \cdot 32 = 32$ …

Why This Approach Works

The Binomial Theorem gives us a systematic way to expand expressions of the form $(a + b)^n$ without having to multiply everything out manually. For $(102)^5$, the trick is to notice that $102$ is very close to $100$ — a nice round number that makes calculations easy. By writing $102 = 100 + 2$, we turn a messy fifth power into a sum of terms that are all multiples of powers of $100$ and $2$.

Here, $a = 100$, $b = 2$, and $n = 5$. Each term in the expansion will be $\binom{5}{k} (100)^{5-k} (2)^k$. Since $100 = 10^2$, powers of $100$ give us nice trailing zeros, and the binomial coefficients are small integers.

Step-by-Step Expansion

1. Write the general expansion

This gives us 6 terms, from $k = 0$ to $k = 5$.

2. Compute the binomial coefficients

The coefficients $\binom{5}{k}$ for $k = 0, 1, 2, 3, 4, 5$ are:

• $\binom{5}{0} = 1$
• $\binom{5}{1} = 5$
• $\binom{5}{2} = 10$
• $\binom{5}{3} = 10$
• $\binom{5}{4} = 5$
• $\binom{5}{5} = 1$

3. Write each term explicitly

For $k = 0$: $\binom{5}{0} (100)^5 (2)^0 = 1 \times 100^5 \times 1 = 100^5$

For $k = 1$: $\binom{5}{1} (100)^4 (2)^1 = 5 \times 100^4 \times 2 = 10 \times 100^4$

For $k = 2$: $\binom{5}{2} (100)^3 (2)^2 = 10 \times 100^3 \times 4 = 40 \times 100^3$

For $k = 3$: $\binom{5}{3} (100)^2 (2)^3 = 10 \times 100^2 \times 8 = 80 \times 100^2$

For $k = 4$: $\binom{5}{4} (100)^1 (2)^4 = 5 \times 100 \times 16 = 80 \times 100$

For $k = 5$: $\binom{5}{5} (100)^0 (2)^5 = 1 \times 1 \times 32 = 32$

4. Compute the powers of 100

• $100^5 = (10^2)^5 = 10^{10} = 10000000000$ (1 followed by 10 zeros)
• $100^4 = 10^8 = 100000000$
• $100^3 = 10^6 = 1000000$
• $100^2 = 10^4 = 10000$
• $100^1 = 100$

5. Multiply each coefficient by its power of 100

• $k = 0$: $1 \times 10000000000 = 10000000000$
• $k = 1$: $10 \times 100000000 = 1000000000$
• $k = 2$: $40 \times 1000000 = 40000000$
• $k = 3$: $80 \times 10000 = 800000$
• $k = 4$: $80 \times 100 = 8000$ …