Compute (98)⁵.

Worked Examples · Example 2

(98)^5

Telangana TsbieTextbookSubjective· 3mImportance★★★★★est

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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}

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
$24!$
›Reveal solutionSolution
The number of odd factors of $24!$ is found by counting divisors formed only from the odd prime factors of $24!$ (ignoring the power of 2). Since $24!$ contains primes up to 23, the odd primes are 3, 5, 7, 11, 13, 17, 19, 23. Multiplying (exponent + 1) for each odd prime gives 10560, so the correct option is (C).

Concept and intuition:

A factor of a number is odd if and only if it contains no factor of 2. So to count the odd factors of $24!$ , we ignore the power of 2 in its prime factorization and consider only the odd primes. The number of divisors of $n = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}$ is $(a_1+1)(a_2+1)\cdots(a_k+1)$ ; for odd factors, we restrict this product to the odd primes only.

Step-by-step solution:

Primes up to 24: 2, 3, 5, 7, 11, 13, 17, 19, 23. We need the exponent of each odd prime in $24!$ , using Legendre's formula $\sum_{i\ge1} \lfloor 24/p^i \rfloor$ .

Exponents of the odd primes in $24!$ :

$p=3$ : $\lfloor 24/3\rfloor + \lfloor 24/9\rfloor = 8+2 = 10$ .

$p=5$ : $\lfloor 24/5\rfloor = 4$ .

$p=7$ : $\lfloor 24/7\rfloor = 3$ .

$p=11$ : $\lfloor 24/11\rfloor = 2$ .

$p=13$ : $\lfloor 24/13\rfloor = 1$ .

$p=17$ : $\lfloor 24/17\rfloor = 1$ .

$p=19$ : $\lfloor 24/19\rfloor = 1$ .

$p=23$ : $\lfloor 24/23\rfloor = 1$ .

Number of odd factors.

For each odd prime, the exponent used in a divisor can range from $0$ up to its exponent in $24!$ , giving (exponent+1) choices. Multiplying across all eight odd primes (four of which — 13, 17, 19, 23 — have exponent 1):

$(10+1)(4+1)(3+1)(2+1)(1+1)^4 = 11 \times 5 \times 4 \times 3 \times 2^4 = 11 \times 5 \times 4 \times 3 \times 16.$
Computing stepwise: $11\times5=55$ ; $55\times4=220$ ; $220\times3=660$ ; $660\times16=10560$ .

Thus the number of odd factors of $24!$ is $10560$ .

Tip

A common mistake is to include the exponent of 2 in the product, which would give the total number of factors (odd and even together). Here we deliberately omit the power of 2.

Watch out

Don't forget that 1 counts as an odd factor (all exponents zero) — it's already included in the product above.

✓Final answer
$The correct option is (C) . ANSWER: C$
TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
$5^{2n} - 48n + k$
›Reveal solutionSolution
The expression must be divisible by 24 for all natural numbers n. Using modular arithmetic and checking n=1 gives k ≡ 23 (mod 24), and verifying with n=2 confirms k=23 works. The least positive integral value of k is 23.

We need the expression $5^{2n} - 48n + k$ to be divisible by 24 for every natural number $n$ . That means for $n = 1, 2, 3, \dots$ , the whole thing leaves remainder 0 when divided by 24.

The key insight: if something must hold for all $n$ , it certainly must hold for the smallest values. We can use those to pin down $k$ . Also, note that $48n$ is already a multiple of 24 (since $48 = 2 \times 24$ ), so it contributes nothing to divisibility by 24 — it’s just along for the ride. The real action is in $5^{2n} + k$ .

Let’s work through it.

Simplify the problem Since $48n$ is always divisible by 24, we can ignore it for divisibility purposes. The condition becomes:

$5^{2n} + k \equiv 0 \pmod{24} \quad \text{for all } n \in \mathbb{N}.$
So we need $5^{2n} + k$ to be a multiple of 24 for every $n$ .

Check the smallest case, $n = 1$ For $n = 1$ :

$5^{2} + k = 25 + k \equiv 0 \pmod{24}.$
Since $25 \equiv 1 \pmod{24}$ , we get:
$1 + k \equiv 0 \pmod{24} \quad \Rightarrow \quad k \equiv 23 \pmod{24}.$
So the smallest positive $k$ that could work is $k = 23$ .

Verify that $k = 23$ works for all $n$ We need to check: does $5^{2n} + 23$ always leave remainder 0 when divided by 24? Notice $5^{2n} = (5^2)^n = 25^n$ . And $25 \equiv 1 \pmod{24}$ , so:

$25^n \equiv 1^n \equiv 1 \pmod{24}.$
Therefore:
$5^{2n} + 23 \equiv 1 + 23 = 24 \equiv 0 \pmod{24}.$
Yes — it works for every $n$ .

Confirm with $n = 2$ as a sanity check For $n = 2$ : $5^4 = 625$ . Divide by 24: $24 \times 26 = 624$ , remainder 1. Then $625 + 23 = 648$ , and $648 / 24 = 27$ , exactly. Works perfectly.

Watch out

A common mistake is to forget that $48n$ is already divisible by 24, so it doesn’t affect the condition. Some students try to expand $5^{2n}$ as $25^n$ but then incorrectly think $25^n$ mod 24 changes with $n$ — it doesn’t, because $25 \equiv 1$ mod 24, so every power is just 1.

Tip

Whenever a divisibility condition must hold for all natural numbers, testing the first few values (especially $n=1$ ) is often the fastest way to find constraints on unknown constants.

✓Final answer
$The correct option is (D) . ANSWER: D$
TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
$p$
›Reveal solutionSolution
The 7th term of the binomial expansion is isolated using the general term formula, set equal to 700, and simplified to find a relation between $p^2$ and $q^2$ , yielding $49p^2 = 9q^2$ .

We are given the expansion $\left(\frac{5}{p^3} - \frac{3q}{7}\right)^8$ and told that the 7th term equals 700. The goal is to find which option correctly relates $49p^2$ to $q^2$ .

Concept & Intuition

In a binomial expansion $(a + b)^n$ , the $(r+1)$ -th term is $T_{r+1} = \binom{n}{r} a^{n-r} b^r$ . Here the second term is negative, so we must be careful with signs. The 7th term corresponds to $r = 6$ (since $r$ starts at 0). We set that term equal to 700 and simplify to find a clean relation between $p^2$ and $q^2$ .

Step-by-step solution

Identify the general term For $\left(\frac{5}{p^3} - \frac{3q}{7}\right)^8$ , let

$a = \frac{5}{p^3}, \quad b = -\frac{3q}{7}, \quad n = 8.$
The $(r+1)$ -th term is
$T_{r+1} = \binom{8}{r} \left(\frac{5}{p^3}\right)^{8-r} \left(-\frac{3q}{7}\right)^r.$

Set $r = 6$ for the 7th term

$T_7 = \binom{8}{6} \left(\frac{5}{p^3}\right)^{2} \left(-\frac{3q}{7}\right)^6.$
Note: $\binom{8}{6} = \binom{8}{2} = 28$ .

Simplify the powers

$\left(\frac{5}{p^3}\right)^2 = \frac{25}{p^6}, \quad \left(-\frac{3q}{7}\right)^6 = \left(\frac{3q}{7}\right)^6 \quad \text{(since even exponent removes the minus sign)}.$
So
$T_7 = 28 \cdot \frac{25}{p^6} \cdot \frac{3^6 q^6}{7^6}.$

Compute the numerical factor $3^6 = 729$ , $7^6 = 117649$ . So

$T_7 = 28 \cdot 25 \cdot \frac{729}{117649} \cdot \frac{q^6}{p^6}.$
Multiply $28 \times 25 = 700$ . Hence
$T_7 = 700 \cdot \frac{729}{117649} \cdot \frac{q^6}{p^6}.$

Set equal to 700

$700 \cdot \frac{729}{117649} \cdot \frac{q^6}{p^6} = 700.$
Cancel 700 (non-zero) to get
$\frac{729}{117649} \cdot \frac{q^6}{p^6} = 1.$

Solve for the ratio

$\frac{q^6}{p^6} = \frac{117649}{729}.$
Notice $117649 = 7^6$ and $729 = 3^6$ . So
$\frac{q^6}{p^6} = \frac{7^6}{3^6} = \left(\frac{7}{3}\right)^6.$
Taking the sixth root (real numbers, so positive root) gives
$\frac{q}{p} = \frac{7}{3} \quad \Rightarrow \quad q = \frac{7}{3}p.$

Find $49p^2$ in terms of $q^2$ From $q = \frac{7}{3}p$ , square both sides:

$q^2 = \frac{49}{9} p^2 \quad \Rightarrow \quad 49p^2 = 9q^2.$
Watch out

A common mistake is to forget that the 7th term uses $r=6$ , not $r=7$ . Also, the negative sign disappears because the exponent is even — but if the exponent were odd, the sign would matter.

Tip

Recognizing $729 = 3^6$ and $117649 = 7^6$ immediately simplifies the ratio without heavy arithmetic.

✓Final answer
$The correct option is (B) . ANSWER: B$

←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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