Binomial Theorem For Any Index

MathsBinomial Theorem And Mathematical InductionFor JEE aspirants

The Binomial Theorem for any index extends the positive-integer expansion of $(1 + x)^{n}$ to arbitrary rational or negative $n$ , at the cost of turning a finite sum into an infinite series that converges only for $∣ x ∣ < 1$ . For JEE aspirants (JEE Main and JEE Advanced), this is the tool of choice for approximating quantities like $99$ , $(1.01)^{- 3}$ , or coefficients in expansions of $(1 - x)^{- n}$ . The general term is $T_{r + 1} = \frac{n ( n - 1 ) \dots ( n - r + 1 )}{r !} x^{r}$ . Alongside this, the multinomial expansion generalises the theorem to sums of more than two terms.

Key Formulas - Quick Reference

General expansion (valid for $∣ x ∣ < 1$ ): $(1 + x)^{n} = 1 + n x + \frac{n ( n - 1 )}{2 !} x^{2} + \frac{n ( n - 1 ) ( n - 2 )}{3 !} x^{3} + \dots$
General term of $(1 + x)^{- n}$ : $T_{r + 1} = (- 1)^{r} \frac{n ( n + 1 ) \dots ( n + r - 1 )}{r !} x^{r}$
General term of $(1 - x)^{- n}$ : $T_{r + 1} = \frac{n ( n + 1 ) \dots ( n + r - 1 )}{r !} x^{r}$
Coefficient of $x^{r}$ in $(1 - x)^{- n}$ : $^{n + r - 1} C_{r}$
$(1 - x)^{- 1} = 1 + x + x^{2} + x^{3} + \dots$ (geometric)
$(1 - x)^{- 2} = 1 + 2 x + 3 x^{2} + 4 x^{3} + \dots$
$(1 - x)^{- 3} = 1 + 3 x + 6 x^{2} + 10 x^{3} + \dots$
Multinomial: total terms in $(x_{1} + x_{2} + \dots + x_{n})^{m} =^{m + n - 1} C_{n - 1}$

1. Binomial Theorem for Any Index

For any real (or complex) exponent $n$ and $∣ x ∣ < 1$ :

(1 + x)^{n} = 1 + n x + \frac{n ( n - 1 )}{2 !} x^{2} + \frac{n ( n - 1 ) ( n - 2 )}{3 !} x^{3} + \dots + \frac{n ( n - 1 ) \dots ( n - r + 1 )}{r !} x^{r} + \dots (to \infty)

Key observations

The expansion is an infinite series when $n$ is not a positive integer; it converges only for $∣ x ∣ < 1$ .
General term of $(1 + x)^{- n}$ (with $n > 0$ ): $T_{r + 1} = (- 1)^{r} \frac{n ( n + 1 ) ( n + 2 ) \dots ( n + r - 1 )}{r !} x^{r}$ .
General term of $(1 - x)^{- n}$ : $T_{r + 1} = \frac{n ( n + 1 ) ( n + 2 ) \dots ( n + r - 1 )}{r !} x^{r}$ (all terms positive).
If the first term is not $1$ , factor it out to reduce to $(1 + y)^{n}$ form: $(a + x)^{n} = a^{n} (1 + \frac{x}{a})^{n} valid for \frac{x}{a} < 1.$

2. Important Standard Expansions

The following six expansions occur repeatedly in JEE problems. Learn them by heart.

Expression	Expansion (valid for $∣ x ∣ < 1$ )
$(1 + x)^{- 1}$	$1 - x + x^{2} - x^{3} + \dots + (- 1)^{r} x^{r} + \dots$
$(1 - x)^{- 1}$	$1 + x + x^{2} + x^{3} + \dots + x^{r} + \dots$
$(1 + x)^{- 2}$	$1 - 2 x + 3 x^{2} - 4 x^{3} + \dots + (- 1)^{r} (r + 1) x^{r} + \dots$
$(1 - x)^{- 2}$	$1 + 2 x + 3 x^{2} + 4 x^{3} + \dots + (r + 1) x^{r} + \dots$
$(1 + x)^{- 3}$	$1 - 3 x + 6 x^{2} - 10 x^{3} + \dots + (- 1)^{r} \frac{( r + 1 ) ( r + 2 )}{2 !} x^{r} + \dots$
$(1 - x)^{- 3}$	$1 + 3 x + 6 x^{2} + 10 x^{3} + \dots + \frac{( r + 1 ) ( r + 2 )}{2 !} x^{r} + \dots$

In general, the coefficient of $x^{r}$ in $(1 - x)^{- n}$ is $^{n + r - 1} C_{r}$ .

Fractional-index expansions

For rational $p / q$ , the expansions are similar with the general term replaced by $\frac{( p / q ) ( p / q - 1 ) \dots ( p / q - r + 1 )}{r !} x^{r}$ . Two useful specialisations:

(1 + x)^{p / q} = 1 + \frac{p}{1 !} (\frac{x}{q}) + \frac{p ( p - q )}{2 !} (\frac{x}{q})^{2} + \dots

(1 - x)^{- p / q} = 1 + \frac{p}{1 !} (\frac{x}{q}) + \frac{p ( p + q )}{2 !} (\frac{x}{q})^{2} + \dots

Solved Example 1

- 1 < x < 1

, show that

(1 - x)^{- 2} = 1 + 2 x + 3 x^{2} + 4 x^{3} + \dots

(to infinity).

Solution:

The general binomial series for any index states, provided $- 1 < x < 1$ ,

$(1 + x)^{n} = 1 + \frac{n}{1 !} x + \frac{n ( n - 1 )}{2 !} x^{2} + \frac{n ( n - 1 ) ( n - 2 )}{3 !} x^{3} + \dots to \infty.$

Substitute $n = - 2$ and replace $x$ with $- x$ :

$(1 - x)^{- 2} = 1 + \frac{- 2}{1 !} (- x) + \frac{( - 2 ) ( - 3 )}{2 !} (- x)^{2} + \frac{( - 2 ) ( - 3 ) ( - 4 )}{3 !} (- x)^{3} + \dots$

$= 1 + 2 x + 3 x^{2} + 4 x^{3} + \dots to \infty.$

Solved Example 2

Find

99

correct to

4

decimal places.

Solution:

Write $99 = 100 - 1 = 100 \cdot (1 - 0.01) = 10 (1 - 0.01)^{1/2}$ .

Apply the binomial series with $n = 1/2$ and small $x = - 0.01$ :

$(1 - 0.01)^{1/2} = 1 + \frac{1/2}{1 !} (- 0.01) + \frac{( 1/2 ) ( - 1/2 )}{2 !} (- 0.01)^{2} + \dots$

$= 1 - 0.005 - 0.0000125 - \dots \approx 0.9949875.$

Therefore $99 \approx 10 \times 0.9949875 = 9.9498 \overline{75} \approx 9.9499$ .

3. Multinomial Expansion

The binomial theorem extends to sums of more than two terms. For positive integers $m, n$ and independent variables $x_{1}, x_{2}, \dots, x_{n}$ :

(x_{1} + x_{2} + \dots + x_{n})^{m} = \sum \frac{m !}{r _{1} ! r _{2} ! \dots r _{n} !} x_{1}^{r_{1}} x_{2}^{r_{2}} \dots x_{n}^{r_{n}}

where the sum is over all

(r_{1}, \dots, r_{n})

with

r_{i} \geq 0

and

r_{1} + r_{2} + \dots + r_{n} = m

Total number of terms in $(x_{1} + x_{2} + \dots + x_{n})^{m}$ is $^{m + n - 1} C_{n - 1}$ .
Coefficient of $x_{1}^{r_{1}} x_{2}^{r_{2}} \dots x_{n}^{r_{n}}$ (with $r_{1} + r_{2} + \dots + r_{n} = m$ ) is $\frac{m !}{r _{1} ! r _{2} ! \dots r _{n} !}$ .
Sum of all coefficients: put $x_{1} = x_{2} = \dots = x_{n} = 1$ , giving $n^{m}$ .

Solved Example 3

x_{1} + x_{2} + x_{3} + x_{4} + x_{5} = 20

and

x_{1} + x_{2} = 5

(with all

x_{i} \geq 0

integers), find the number of non-negative integer solutions.

Solution:

The system splits into two independent problems:

(a) $x_{1} + x_{2} = 5$ : number of non-negative integer solutions $=^{5 + 2 - 1} C_{2 - 1} =^{6} C_{1} = 6$ .

(b) $x_{3} + x_{4} + x_{5} = 15$ : number of solutions $=^{15 + 3 - 1} C_{3 - 1} =^{17} C_{2} = 136$ .

Total solutions $= 6 \times 136 = 816$ .

Equivalent form. The number of non-negative integer solutions of $y_{1} + y_{2} + \dots + y_{k} = N$ equals the coefficient of $x^{N}$ in $(1 - x)^{- k}$ , which is $^{N + k - 1} C_{k - 1}$ .

Common Mistakes to Avoid

Watch out

Applying the any-index expansion for $∣ x ∣ \geq 1$ : the series diverges. Always check the validity region first.
Forgetting to factor out to bring the binomial into $(1 + y)^{n}$ form before expanding: e.g., $(2 + x)^{- 3}$ becomes $2^{- 3} (1 + x /2)^{- 3}$ , valid for $∣ x /2∣ < 1$ i.e. $∣ x ∣ < 2$ .
Confusing the signs: $(1 + x)^{- n}$ has alternating signs, but $(1 - x)^{- n}$ has all positive signs.
In multinomial expansion, mistaking "total terms" for "distinct powers of a variable" - they are different.
Truncating the approximation too early: for $4$ decimal accuracy, keep terms until the next term is smaller than $5 \times 1 0^{- 5}$ .

Frequently Asked Questions

What is the Binomial Theorem for any index and when is it valid?

The Binomial Theorem for any index generalises $(1 + x)^{n}$ from positive integer $n$ to any rational (or real) $n$ . The expansion $(1 + x)^{n} = 1 + n x + \frac{n ( n - 1 )}{2 !} x^{2} + \dots$ becomes an infinite series and is valid only for $∣ x ∣ < 1$ . This is a Class 11 topic in the JEE Main and JEE Advanced syllabus; NEET does not include Mathematics.

Why does the binomial series for negative or fractional index have infinitely many terms?

When $n$ is a positive integer, one of the factors $(n - r)$ in the general term $\frac{n ( n - 1 ) \dots ( n - r + 1 )}{r !}$ becomes zero for $r > n$ , terminating the series. For negative or fractional $n$ , no factor is ever zero, so the series continues indefinitely. It converges only for $∣ x ∣ < 1$ .

What is the expansion of $(1 - x)^{- n}$ where $n$ is a positive integer?

$(1 - x)^{- n} = 1 + n x + \frac{n ( n + 1 )}{2 !} x^{2} + \frac{n ( n + 1 ) ( n + 2 )}{3 !} x^{3} + \dots$ , with general term $T_{r + 1} = \frac{n ( n + 1 ) \dots ( n + r - 1 )}{r !} x^{r}$ . Equivalently, the coefficient of $x^{r}$ is $^{n + r - 1} C_{r}$ . All terms are positive. Valid for $∣ x ∣ < 1$ .

How do I use the binomial theorem to approximate $99$ or $(1.01)^{- 3}$ ?

Write the number in the form $a (1 + y)^{n}$ with $∣ y ∣$ small. Example: $99 = 10 (1 - 0.01)^{1/2}$ . Expand using the binomial series, keep enough terms so the next term is smaller than the required accuracy, then multiply back by $a$ . For 4-decimal accuracy, keep terms until $∣ T_{r + 1} ∣ < 5 \times 1 0^{- 5}$ .

What is the multinomial theorem?

The multinomial theorem generalises the binomial theorem to a sum of more than two terms: $(x_{1} + x_{2} + \dots + x_{n})^{m} = \sum \frac{m !}{r _{1} ! r _{2} ! \dots r _{n} !} x_{1}^{r_{1}} \dots x_{n}^{r_{n}}$ , summed over all $r_{i} \geq 0$ with $r_{1} + \dots + r_{n} = m$ . The number of terms is $^{m + n - 1} C_{n - 1}$ .

How many terms are there in the expansion of $(x_{1} + x_{2} + \dots + x_{n})^{m}$ ?

The total number of terms equals $^{m + n - 1} C_{n - 1}$ , which is the number of non-negative integer solutions $(r_{1}, r_{2}, \dots, r_{n})$ of $r_{1} + r_{2} + \dots + r_{n} = m$ . For example, $(x + y + z)^{4}$ has $^{6} C_{2} = 15$ distinct terms.

How is the multinomial theorem used to count non-negative integer solutions?

The number of non-negative integer solutions of $y_{1} + y_{2} + \dots + y_{k} = N$ equals the coefficient of $x^{N}$ in $(1 - x)^{- k}$ , which is $^{N + k - 1} C_{k - 1}$ . This links combinatorics ('stars and bars') with the binomial-for-any-index expansion of $(1 - x)^{- k}$ .

What is the relation between coefficient of $x^{r}$ in $(1 - x)^{- n}$ and combinations with repetition?

The coefficient of $x^{r}$ in $(1 - x)^{- n}$ is $^{n + r - 1} C_{r}$ , which also counts the number of ways to choose $r$ objects from $n$ types with repetition allowed. This identity ties the binomial-for-any-index expansion directly to combinatorial 'stars and bars' counting.

Can the any-index binomial theorem be applied when the first term is not 1?

Yes. Factor out the first term: $(a + x)^{n} = a^{n} (1 + \frac{x}{a})^{n}$ , valid whenever $\frac{x}{a} < 1$ , i.e. $∣ x ∣ < ∣ a ∣$ . Then expand using the standard series in $x / a$ .

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1. Binomial Theorem for Any Index

Key observations

2. Important Standard Expansions

Fractional-index expansions

3. Multinomial Expansion

Common Mistakes to Avoid

Frequently Asked Questions

What is the Binomial Theorem for any index and when is it valid?

Why does the binomial series for negative or fractional index have infinitely many terms?

What is the expansion of (1−x)−n where n is a positive integer?

How do I use the binomial theorem to approximate 99​ or (1.01)−3?

What is the multinomial theorem?

How many terms are there in the expansion of (x1​+x2​+⋯+xn​)m?

How is the multinomial theorem used to count non-negative integer solutions?

What is the relation between coefficient of xr in (1−x)−n and combinations with repetition?

Can the any-index binomial theorem be applied when the first term is not 1?

Ready to master Binomial Theorem And Mathematical Induction?

What is the expansion of $(1 - x)^{- n}$ where $n$ is a positive integer?

How do I use the binomial theorem to approximate $99$ or $(1.01)^{- 3}$ ?

How many terms are there in the expansion of $(x_{1} + x_{2} + \dots + x_{n})^{m}$ ?

What is the relation between coefficient of $x^{r}$ in $(1 - x)^{- n}$ and combinations with repetition?