Compute (98)⁵.

Worked Examples · Example 2

(98)^5

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Concept understanding — Binomial Theorem

The Binomial Theorem: From Patterns to Power

Imagine you have to expand $(x + y)^2$ . You know it's $x^2 + 2xy + y^2$ . What about $(x + y)^3$ ? That's $x^3 + 3x^2y + 3xy^2 + y^3$ . Now try $(x + y)^4$ — you could multiply $(x + y)^3$ by $(x + y)$ again, but it gets messy fast.

The Binomial Theorem is the shortcut. It tells you exactly what $(x + y)^n$ expands to, for any positive integer $n$ , without doing the multiplication step by step.

The Pattern You Already Know

Look at the expansions we have:

Power	Expansion
$(x + y)^0$	$1$
$(x + y)^1$	$x + y$
$(x + y)^2$	$x^2 + 2xy + y^2$
$(x + y)^3$	$x^3 + 3x^2y + 3xy^2 + y^3$
$(x + y)^4$	$x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$

Three things stand out:

The powers of $x$ decrease from $n$ down to $0$ , while the powers of $y$ increase from $0$ up to $n$ . In each term, the exponents add to $n$ .
The coefficients — 1, 2, 1 for $n=2$ ; 1, 3, 3, 1 for $n=3$ ; 1, 4, 6, 4, 1 for $n=4$ — follow a famous pattern called Pascal's triangle.
The number of terms is always $n+1$ .

Note

Pascal's triangle: each number is the sum of the two numbers directly above it.

Why Do These Coefficients Appear?

Think about what $(x + y)^n$ really means. It's $(x + y)$ multiplied by itself $n$ times:

(x + y)^n = \underbrace{(x + y)(x + y)\cdots(x + y)}_{n \text{ factors}}

When you expand, you pick either $x$ or $y$ from each factor. A term like $x^{n-k}y^k$ comes from choosing $y$ from exactly $k$ of the $n$ factors and $x$ from the rest.

How many ways can you choose which $k$ factors give you $y$ ? That's exactly the number of combinations: $\binom{n}{k}$ (read "n choose k").

\binom{n}{k} = \frac{n!}{k!(n-k)!}

So the coefficient of $x^{n-k}y^k$ is $\binom{n}{k}$ . That's the heart of the theorem.

The Precise Statement

Important

Binomial Theorem: For any positive integer $n$ ,

$(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k$

Or written out:

$(x + y)^n = \binom{n}{0}x^n + \binom{n}{1}x^{n-1}y + \binom{n}{2}x^{n-2}y^2 + \cdots + \binom{n}{n-1}xy^{n-1} + \binom{n}{n}y^n$

Notice $\binom{n}{0} = 1$ and $\binom{n}{n} = 1$ , which matches the first and last coefficients always being 1.

A Quick Example

Expand $(2a - b)^5$ using the theorem.

Here $x = 2a$ , $y = -b$ , and $n = 5$ .

(2a - b)^5 = \sum_{k=0}^{5} \binom{5}{k} (2a)^{5-k} (-b)^k

Compute term by term:

$k=0$ : $\binom{5}{0}(2a)^5(-b)^0 = 1 \cdot 32a^5 = 32a^5$
$k=1$ : $\binom{5}{1}(2a)^4(-b)^1 = 5 \cdot 16a^4 \cdot (-b) = -80a^4b$
$k=2$ : $\binom{5}{2}(2a)^3(-b)^2 = 10 \cdot 8a^3 \cdot b^2 = 80a^3b^2$
$k=3$ : $\binom{5}{3}(2a)^2(-b)^3 = 10 \cdot 4a^2 \cdot (-b^3) = -40a^2b^3$
$k=4$ : $\binom{5}{4}(2a)^1(-b)^4 = 5 \cdot 2a \cdot b^4 = 10ab^4$
$k=5$ : $\binom{5}{5}(2a)^0(-b)^5 = 1 \cdot 1 \cdot (-b^5) = -b^5$

So:

(2a - b)^5 = 32a^5 - 80a^4b + 80a^3b^2 - 40a^2b^3 + 10ab^4 - b^5

Watch out

A common mistake: forgetting the sign when $y$ is negative. Here $(-b)^k$ alternates signs — even $k$ gives positive, odd $k$ gives negative.

Why This Matters

The Binomial Theorem isn't just for expanding brackets. It appears in probability (binomial distribution), calculus (binomial series for non-integer exponents), and even in estimating powers without a calculator. Once you see the pattern, you'll spot it everywhere.

The key takeaway: every term in $(x + y)^n$ is of the form $\binom{n}{k} x^{n-k} y^k$ . The theorem gives you all $n+1$ terms in one clean formula.

The Binomial Theorem itself, along with Pascal's triangle and the general term formula, is one of the most heavily tested chapters in NCERT Class 11 Mathematics, and "binomial theorem class 11 formula, definition and examples" is a frequently searched revision query for CBSE boards and JEE Main. Because the theorem also underlies probability and approximation problems, it consistently appears in "binomial theorem important questions" compiled for competitive-exam practice.

The key idea is to rewrite $98$ as $(100 - 2)$ and apply the Binomial Theorem. The expansion gives $100^5 - 5 \cdot 100^4 \cdot 2 + 10 \cdot 100^3 \cdot 4 - 10 \cdot 100^2 \cdot 8 + 5 \cdot 100 \cdot 16 - 32$ , which simplifies to $9,\!039,\!207,\!968$ .

Why the Binomial Theorem works here

Directly multiplying $98$ five times is tedious and error-prone. But $98$ is very close to $100$ — a round number that's easy to raise to powers. The Binomial Theorem lets us expand $(a + b)^n$ as a sum of terms, each involving powers of $a$ and $b$ with binomial coefficients. By writing $98 = 100 - 2$ , we turn a messy multiplication into a clean sum of just six terms, each of which is simple to compute.

(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k

Here $a = 100$ , $b = -2$ , and $n = 5$ .

Step-by-step expansion

1. Write the expression in binomial form

(98)^5 = (100 - 2)^5

We'll use $a = 100$ , $b = -2$ , $n = 5$ .

2. Write out the general term

The $k$ -th term (starting from $k=0$ ) is:

\binom{5}{k} (100)^{5-k} (-2)^k

We need terms for $k = 0, 1, 2, 3, 4, 5$ .

3. Compute the binomial coefficients

\binom{5}{0} = 1,\quad \binom{5}{1} = 5,\quad \binom{5}{2} = 10,\quad \binom{5}{3} = 10,\quad \binom{5}{4} = 5,\quad \binom{5}{5} = 1

4. Compute each term carefully

$k = 0$ : $\binom{5}{0} (100)^5 (-2)^0 = 1 \cdot 100^5 \cdot 1 = 10,\!000,\!000,\!000$
$k = 1$ : $\binom{5}{1} (100)^4 (-2)^1 = 5 \cdot 100^4 \cdot (-2)$

$100^4 = 100,\!000,\!000$ , so $5 \times 100,\!000,\!000 = 500,\!000,\!000$ , times $(-2)$ gives $-1,\!000,\!000,\!000$
$k = 2$ : $\binom{5}{2} (100)^3 (-2)^2 = 10 \cdot 100^3 \cdot 4$

$100^3 = 1,\!000,\!000$ , so $10 \times 1,\!000,\!000 = 10,\!000,\!000$ , times $4$ gives $40,\!000,\!000$
$k = 3$ : $\binom{5}{3} (100)^2 (-2)^3 = 10 \cdot 100^2 \cdot (-8)$

$100^2 = 10,\!000$ , so $10 \times 10,\!000 = 100,\!000$ , times $(-8)$ gives $-800,\!000$
$k = 4$ : $\binom{5}{4} (100)^1 (-2)^4 = 5 \cdot 100 \cdot 16$

$5 \times 100 = 500$ , times $16$ gives $8,\!000$
$k = 5$ : $\binom{5}{5} (100)^0 (-2)^5 = 1 \cdot 1 \cdot (-32) = -32$

Watch out

A common mistake is forgetting the sign when $b$ is negative. Here $(-2)^k$ alternates sign: positive for even $k$ , negative for odd $k$ . Double-check each term's sign before adding.

5. Add all terms

10,\!000,\!000,\!000 - 1,\!000,\!000,\!000 = 9,\!000,\!000,\!000

9,\!000,\!000,\!000 + 40,\!000,\!000 = 9,\!040,\!000,\!000

9,\!040,\!000,\!000 - 800,\!000 = 9,\!039,\!200,\!000

9,\!039,\!200,\!000 + 8,\!000 = 9,\!039,\!208,\!000

9,\!039,\!208,\!000 - 32 = 9,\!039,\!207,\!968

Tip

Notice how the terms decrease dramatically in size: the first term is $10$ billion, the last is just $-32$ . The Binomial Theorem lets you handle huge numbers by breaking them into manageable pieces.

✓Final answer

\boxed{9,\!039,\!207,\!968}

←Previous question · 7.2.2Example 1 — Expand (x² + (3)/(x))⁴, x ≠ 0.→Next question · 7.2.2Example 3 — Which is larger (1.01)¹⁰⁰⁰⁰⁰⁰ or 10,000?

Q.Compute (98)5(98)^5(98)5.

Why the Binomial Theorem works here

Step-by-step expansion

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Q.Compute $(98)^5$ .

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