Numerical Applications of the Binomial Theorem

Big Powers by Hand

The theorem computes $(98)^5$ faster than repeated multiplication: write the base as a sum or difference of numbers with easy powers, then expand.

For example, $98 = 100 - 2$, so

$$(98)^5 = {^5C_0}(100)^5 - {^5C_1}(100)^4(2) + {^5C_2}(100)^3(2^2) - {^5C_3}(100)^2(2^3) + {^5C_4}(100)(2^4) - {^5C_5}(2^5)$$

$$= 10000000000 - 1000000000 + 40000000 - 800000 + 8000 - 32 = 9039207968$$

Same recipe: $96 = 100 - 4$, $102 = 100 + 2$, $101 = 100 + 1$, $99 = 100 - 1$, $(1.1) = 1 + 0.1$, $(0.99) = 1 - 0.01$.

Key Point: For a minus split the signs alternate — track them term by term. Powers of 100 shift the decimal, powers of the small number stay tiny; the terms are easy to add.

[Board Tip] For an approximation, keep only the first few terms: $(0.99)^5 = (1 - 0.01)^5 \approx 1 - 5(0.01) + 10(0.01)^2 = 0.951$ — the dropped terms are of order $10^{-6}$.

Comparisons and Divisibility

Which is larger? Expand the bigger-looking side and keep just enough terms:

For example: $(1.01)^{1000000} = (1 + 0.01)^{1000000} = 1 + 1000000(0.01) + \text{positive terms} = 1 + 10000 + \cdots > 10000$.

Key Point: For $x > 0$, every term of $(1+x)^n$ is positive, so $(1+x)^n > 1 + nx$ after just two terms. That single inequality settles all the standard comparisons.

Divisibility and remainders come from splitting the base and pulling out the modulus:

$$6^n = (1+5)^n = 1 + 5n + 5^2\,{^nC_2} + 5^3\,{^nC_3} + \cdots = 1 + 5n + 25\left({^nC_2} + 5\,{^nC_3} + \cdots\right)$$

so $6^n - 5n = 25k + 1$: dividing $6^n - 5n$ by 25 always leaves remainder 1. The same recipe shows more: $9^{n+1} = (1+8)^{n+1} = 1 + 8(n+1) + 64\left({^{n+1}C_2} + \cdots\right)$, so $9^{n+1} - 8n - 9$ is divisible by 64.

[JEE Tip] Choose the split so the modulus appears squared in the third term: to work mod 25 split off 5's; mod 64 split off 8's; mod 49 split off 7's. The first two terms then carry all the remainder information.

Conjugate Pairs: Surds That Cancel

Adding or subtracting the expansions of $(a+b)^n$ and $(a-b)^n$ kills half the terms:

$$(a+b)^n + (a-b)^n = 2\left[{^nC_0}a^n + {^nC_2}a^{n-2}b^2 + \cdots\right] \quad \text{(even powers of } b\text{)}$$

$$(a+b)^n - (a-b)^n = 2\left[{^nC_1}a^{n-1}b + {^nC_3}a^{n-3}b^3 + \cdots\right] \quad \text{(odd powers of } b\text{)}$$

With surd values of $b$, the surviving powers are even, so the surds vanish and the result is often an integer. For example: $(x+1)^6 + (x-1)^6 = 2(x^6 + 15x^4 + 15x^2 + 1)$, and at $x = \sqrt{2}$: $2(8 + 60 + 30 + 1) = 198$.

[Board Tip] Work symbolically first — find the polynomial identity in $x$, and substitute the surd only at the end. Substituting early buries you in surd arithmetic.

Solved Examples

Example 1: $(98)^5$

Compute $(98)^5$.

Solution:

Step 1 — Split the base. $98 = 100 - 2$: expand $(100-2)^5$ with alternating signs.

Step 2 — Write the six terms. $10^{10} - 5(10^8)(2) + 10(10^6)(4) - 10(10^4)(8) + 5(100)(16) - 32$.

Step 3 — Evaluate. $10000000000 - 1000000000 + 40000000 - 800000 + 8000 - 32$.

Step 4 — Add. $= 9039207968$.

Takeaway: Powers of 100 shift the decimal; powers of 2 stay tiny — six easy terms beat four long multiplications.

Example 2: $(96)^3$

Evaluate $(96)^3$.

Solution:

Step 1 — Split. $96 = 100 - 4$: $(100-4)^3 = 10^6 - 3(10^4)(4) + 3(100)(16) - 64$.

Step 2 — Evaluate. $1000000 - 120000 + 4800 - 64$.

Step 3 — Add. $= 884736$.

Takeaway: Row 3 ($1, 3, 3, 1$) and powers of 4 — sign alternation from the minus split.

Example 3: $(102)^5$

Evaluate $(102)^5$.

Solution:

Step 1 — Split. $102 = 100 + 2$ — a plus split, so ALL signs are positive.

Step 2 — Write the terms. $10^{10} + 5(10^8)(2) + 10(10^6)(4) + 10(10^4)(8) + 5(100)(16) + 32$.

Step 3 — Add. $10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32 = 11040808032$.

Takeaway: Same skeleton as $(98)^5$ — only the signs differ between $100-2$ and $100+2$.

Example 4: $(101)^4$

Evaluate $(101)^4$.

Solution:

Step 1 — Split. $(100+1)^4$: powers of 1 vanish, leaving the row itself.

Step 2 — Write the terms. $10^8 + 4(10^6) + 6(10^4) + 4(100) + 1$.

Step 3 — Add. $100000000 + 4000000 + 60000 + 400 + 1 = 104060401$.

Takeaway: The digits of row 4 — $1, 4, 6, 4, 1$ — appear directly in the answer: 1 04 06 04 01.

Example 5: $(99)^5$

Evaluate $(99)^5$.

Solution:

Step 1 — Split. $(100-1)^5$: powers of 1 vanish, signs alternate.

Step 2 — Write the terms. $10^{10} - 5(10^8) + 10(10^6) - 10(10^4) + 5(100) - 1$.

Step 3 — Add. $10000000000 - 500000000 + 10000000 - 100000 + 500 - 1 = 9509900499$.

Takeaway: A $\pm 1$ split is the cleanest case — the row entries are the whole computation.

Example 6: The million-power comparison

Which is larger: $(1.01)^{1000000}$ or $10000$?

Solution:

Step 1 — Split the base. $(1.01)^{1000000} = (1 + 0.01)^{1000000}$.

Step 2 — Keep the first two terms. $= 1 + 1000000(0.01) + \text{other positive terms} = 1 + 10000 + \cdots$

Step 3 — Compare. $1 + 10000 + \text{positive terms} > 10000$. So $(1.01)^{1000000}$ is larger.

Takeaway: For $x > 0$, $(1+x)^n > 1 + nx$ — two terms of the expansion settle the comparison.

Example 7: A second comparison

Which is larger: $(1.1)^{10000}$ or $1000$?

Solution:

Step 1 — Two-term bound. $(1 + 0.1)^{10000} > 1 + 10000(0.1) = 1001$.

Step 2 — Compare. $1001 > 1000$, so $(1.1)^{10000}$ is larger.

Takeaway: The same inequality works whenever the truncated value already beats the target.

Example 8: Approximation

Find an approximation of $(0.99)^5$ using the first three terms of its expansion.

Solution:

Step 1 — Split. $(0.99)^5 = (1 - 0.01)^5$.

Step 2 — Keep three terms. $1 - 5(0.01) + 10(0.01)^2 = 1 - 0.05 + 0.001$.

Step 3 — Evaluate. $\approx 0.951$. The dropped terms are of order $10(0.01)^3 = 10^{-5}$ — negligible at three decimals.

Takeaway: Approximations keep the first few terms and estimate the size of the first dropped one.

Example 9: Remainder mod 25

Prove that $6^n - 5n$ always leaves remainder 1 when divided by 25.

Solution:

Step 1 — Split the base. $6^n = (1+5)^n = 1 + 5n + 5^2\,{^nC_2} + 5^3\,{^nC_3} + \cdots + 5^n$.

Step 2 — Group the 25's. From the third term onward every term has factor $5^2 = 25$:

$$6^n - 5n = 1 + 25\left({^nC_2} + 5\,{^nC_3} + \cdots + 5^{n-2}\right) = 25k + 1$$

Step 3 — Conclude. The remainder on division by 25 is always 1. ∎

Takeaway: Split so the modulus appears squared in the third term — the first two terms carry the remainder.

Example 10: Divisible by 64

Show that $9^{n+1} - 8n - 9$ is divisible by 64 for every positive integer $n$.

Solution:

Step 1 — Split. $9^{n+1} = (1+8)^{n+1} = 1 + 8(n+1) + 8^2\,{^{n+1}C_2} + 8^3\,{^{n+1}C_3} + \cdots$

Step 2 — Subtract the linear part. $1 + 8(n+1) = 8n + 9$, so

$$9^{n+1} - 8n - 9 = 64\left({^{n+1}C_2} + 8\,{^{n+1}C_3} + \cdots\right)$$

Step 3 — Conclude. The bracket is an integer, so the expression is a multiple of 64. ∎

Takeaway: Same recipe as mod 25 — this time the 8's supply $8^2 = 64$.

Example 11: A conjugate difference

Find $(a+b)^4 - (a-b)^4$, and evaluate $(\sqrt{3} + \sqrt{2})^4 - (\sqrt{3} - \sqrt{2})^4$.

Solution:

Step 1 — Subtract the twin expansions. Even powers of $b$ cancel; odd powers double:

$$(a+b)^4 - (a-b)^4 = 2\left[{^4C_1}a^3b + {^4C_3}ab^3\right] = 2\left[4a^3b + 4ab^3\right] = 8ab(a^2 + b^2)$$

Step 2 — Substitute the surds. $a = \sqrt{3}, b = \sqrt{2}$: $ab = \sqrt{6}$ and $a^2 + b^2 = 3 + 2 = 5$.

Step 3 — Evaluate. $8\sqrt{6} \times 5 = 40\sqrt{6}$.

Takeaway: Find the polynomial identity first; substitute surds last.

Example 12: A conjugate sum

Find $(x+1)^6 + (x-1)^6$, and evaluate $(\sqrt{2}+1)^6 + (\sqrt{2}-1)^6$.

Solution:

Step 1 — Add the twin expansions. Odd powers cancel; even powers double:

$$(x+1)^6 + (x-1)^6 = 2\left[x^6 + 15x^4 + 15x^2 + 1\right]$$

Step 2 — Substitute $x = \sqrt{2}$. $x^2 = 2$, $x^4 = 4$, $x^6 = 8$: $2(8 + 60 + 30 + 1)$.

Step 3 — Evaluate. $2 \times 99 = 198$ — an integer; the surds vanished with the odd powers.

Takeaway: Conjugate sums keep even powers, where $(\sqrt{2})^{\text{even}}$ is rational.

Example 13: A factor theorem via expansion

If $a$ and $b$ are distinct integers, prove $a - b$ is a factor of $a^n - b^n$.

Solution:

Step 1 — Rewrite the base. $a = (a - b) + b$, so $a^n = \left[(a-b) + b\right]^n$.

Step 2 — Expand with first quantity $(a-b)$.

$$a^n = (a-b)^n + {^nC_1}(a-b)^{n-1}b + \cdots + {^nC_{n-1}}(a-b)b^{n-1} + b^n$$

Step 3 — Subtract $b^n$ and factor. Every remaining term contains $(a-b)$:

$$a^n - b^n = (a-b)\left[(a-b)^{n-1} + {^nC_1}(a-b)^{n-2}b + \cdots + {^nC_{n-1}}b^{n-1}\right]$$

The bracket is an integer, so $(a-b) \mid a^n - b^n$. ∎

Takeaway: Adding-and-subtracting inside the base ($a = (a-b) + b$) is the theorem's sneakiest, most reusable move.