Properties Of Binomial Coefficient

MathsBinomial Theorem And Mathematical InductionFor JEE aspirants

The binomial coefficients $^{n} C_{0},^{n} C_{1}, \dots,^{n} C_{n}$ in the expansion of $(1 + x)^{n}$ satisfy a rich set of identities. For JEE Mathematics (JEE Main and JEE Advanced), these properties power a large class of series-summation problems. Key identities include $\sum C_{r} = 2^{n}$ , $\sum (- 1)^{r} C_{r} = 0$ , and the Pascal recurrence $^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$ . Standard techniques - substitution, differentiation, integration, and multiplication of expansions - convert any coefficient series into a closed form. Mastering these tools is essential for high-yield JEE Advanced problems.

Key Formulas - Quick Reference

Sum of all coefficients: $C_{0} + C_{1} + C_{2} + \dots + C_{n} = 2^{n}$
Alternating sum: $C_{0} - C_{1} + C_{2} - \dots + (- 1)^{n} C_{n} = 0$
Even-index sum = Odd-index sum: $C_{0} + C_{2} + C_{4} + \dots = C_{1} + C_{3} + \dots = 2^{n - 1}$
Symmetry: $^{n} C_{r_{1}} =^{n} C_{r_{2}} \Rightarrow r_{1} = r_{2}$ or $r_{1} + r_{2} = n$
Pascal's rule: $^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$
Factor-of- $r$ identity: $r \cdot^{n} C_{r} = n \cdot^{n - 1} C_{r - 1}$
Divide-by- $(r + 1)$ : $\frac{^{n} C _{r}}{r + 1} = \frac{^{n + 1} C _{r + 1}}{n + 1}$
Sum with $r$ : $\sum_{r = 0}^{n} r^{n} C_{r} = n \cdot 2^{n - 1}$

1. Standard Notation and Basic Properties

For convenience, the coefficients $^{n} C_{0},^{n} C_{1}, \dots,^{n} C_{n}$ appearing in the expansion of $(1 + x)^{n}$ are usually written simply as $C_{0}, C_{1}, \dots, C_{n}$ (the value of $n$ being understood from context). The most useful identities are the following.

$C_{0} + C_{1} + C_{2} + \dots + C_{n} = 2^{n}$ (put $x = 1$ in $(1 + x)^{n}$ ).
$C_{0} - C_{1} + C_{2} - \dots + (- 1)^{n} C_{n} = 0$ (put $x = - 1$ ).
$C_{0} + C_{2} + C_{4} + \dots = C_{1} + C_{3} + C_{5} + \dots = 2^{n - 1}$ (add and subtract the two identities above).
$^{n} C_{r_{1}} =^{n} C_{r_{2}} \Rightarrow r_{1} = r_{2}$ or $r_{1} + r_{2} = n$ (symmetry of Pascal's row).
$^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$ (Pascal's rule).
$r \cdot^{n} C_{r} = n \cdot^{n - 1} C_{r - 1}$ (useful for series with a factor of $r$ ).
$\frac{^{n} C _{r}}{r + 1} = \frac{^{n + 1} C _{r + 1}}{n + 1}$ (useful for series with a factor of $\frac{1}{r + 1}$ ).

Solved Example 1

Find the value of

r = 0 \sum n (\frac{r + 2}{r + 1})^{n} C_{r}

Solution:

Write $\frac{r + 2}{r + 1} = 1 + \frac{1}{r + 1}$ . Then

$r = 0 \sum n (\frac{r + 2}{r + 1})^{n} C_{r} = r = 0 \sum n^{n} C_{r} + r = 0 \sum n \frac{^{n} C _{r}}{r + 1}$

The first sum is $2^{n}$ . For the second, use $\frac{^{n} C _{r}}{r + 1} = \frac{^{n + 1} C _{r + 1}}{n + 1}$ :

$r = 0 \sum n \frac{^{n} C _{r}}{r + 1} = \frac{1}{n + 1} r = 0 \sum n^{n + 1} C_{r + 1} = \frac{1}{n + 1} (2^{n + 1} - 1)$

Total: $2^{n} + \frac{2 ^{n + 1} - 1}{n + 1} = \frac{2 ^{n} ( n + 3 ) - 1}{n + 1}$ .

Solved Example 2

(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}

, show that

(C_{0} + C_{1}) (C_{1} + C_{2}) (C_{2} + C_{3}) \dots (C_{n - 1} + C_{n}) = \frac{( n + 1 ) ^{n}}{n !} C_{1} C_{2} \dots C_{n}

Solution:

By Pascal's rule, $C_{r - 1} + C_{r} =^{n + 1} C_{r} = \frac{( n + 1 )!}{r ! ( n + 1 - r )!} = \frac{n + 1}{r} \cdot \frac{n !}{( r - 1 )! ( n - r + 1 )!} = \frac{n + 1}{r} C_{r - 1}$ .

Applying this for $r = 1, 2, \dots, n$ :

$C_{0} + C_{1} = \frac{n + 1}{1} C_{0}$ , $C_{1} + C_{2} = \frac{n + 1}{2} C_{1}$ , $\dots$ , $C_{n - 1} + C_{n} = \frac{n + 1}{n} C_{n - 1}$ .

Multiplying:

$(C_{0} + C_{1}) (C_{1} + C_{2}) \dots (C_{n - 1} + C_{n}) = \frac{( n + 1 ) ^{n}}{n !} C_{0} C_{1} \dots C_{n - 1}$

Since $C_{0} = 1$ and by symmetry $C_{k} = C_{n - k}$ , we have $C_{0} C_{1} \dots C_{n - 1} = C_{1} C_{2} \dots C_{n}$ . Hence the required identity holds.

2. Series with Alternating Signs

Series of binomial coefficients with alternating signs are handled by substituting $x = - 1$ in $(1 + x)^{n}$ (and, when needed, cube roots of unity to isolate every third coefficient).

Solved Example 3

Evaluate

C_{0} - C_{1} + C_{2} - C_{3} + \dots + (- 1)^{n} C_{n}

Solution:

Substitute $x = - 1$ in $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}$ :

$(1 - 1)^{n} = 0 = C_{0} - C_{1} + C_{2} - \dots + (- 1)^{n} C_{n}$

So the alternating sum equals $0$ .

Related. Adding $(1 + 1)^{n}$ and $(1 - 1)^{n}$ gives $C_{0} + C_{2} + C_{4} + \dots = 2^{n - 1}$ . To isolate $C_{0} + C_{3} + C_{6} + \dots$ , substitute $x = 1, ω, ω^{2}$ (cube roots of unity) and add.

3. Series with a Factor of an Integer (Differentiation Trick)

When each term of a series has the form $k^{n} C_{r}$ (with $k$ a linear function of $r$ ), differentiate $(1 + x)^{n}$ and then substitute a suitable value of $x$ .

Solved Example 4

(1 + x)^{n} = r = 0 \sum n C_{r} x^{r}

, prove that

C_{1} + 2 C_{2} + 3 C_{3} + \dots + n C_{n} = n \cdot 2^{n - 1}

Solution:

Method 1 (summation). Use $r \cdot^{n} C_{r} = n \cdot^{n - 1} C_{r - 1}$ :

$r = 1 \sum n r^{n} C_{r} = n r = 1 \sum n^{n - 1} C_{r - 1} = n (^{n - 1} C_{0} +^{n - 1} C_{1} + \dots +^{n - 1} C_{n - 1}) = n \cdot 2^{n - 1} .$

Method 2 (calculus). Differentiate $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}$ with respect to $x$ :

$n (1 + x)^{n - 1} = C_{1} + 2 C_{2} x + 3 C_{3} x^{2} + \dots + n C_{n} x^{n - 1}$

Put $x = 1$ : $n \cdot 2^{n - 1} = C_{1} + 2 C_{2} + 3 C_{3} + \dots + n C_{n}$ .

4. Series with a Coefficient Divided by an Integer (Integration Trick)

When each term of a series has the form $\frac{^{n} C _{r}}{k}$ (with $k$ linear in $r$ ), integrate $(1 + x)^{n}$ within suitable limits.

Solved Example 5

Prove that

\frac{C _{0}}{1} + \frac{C _{2}}{3} + \frac{C _{4}}{5} + \dots = \frac{2 ^{n}}{n + 1}

Solution:

Start with $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + C_{3} x^{3} + \dots + C_{n} x^{n}$ . Integrate both sides from $- 1$ to $1$ :

$\int_{- 1}^{1} (1 + x)^{n} d x = [\frac{( 1 + x ) ^{n + 1}}{n + 1}]_{- 1}^{1} = \frac{2 ^{n + 1}}{n + 1} .$

On the right, the odd-power terms integrate to $0$ (odd function over a symmetric interval), leaving only the even-power terms:

$\int_{- 1}^{1} (1 + x)^{n} d x = 2 (C_{0} + \frac{C _{2}}{3} + \frac{C _{4}}{5} + \dots) .$

Therefore $C_{0} + \frac{C _{2}}{3} + \frac{C _{4}}{5} + \dots = \frac{2 ^{n}}{n + 1}$ .

5. Series with Product of Two Binomial Coefficients

Series where each term is a product of two coefficients - e.g., $C_{1} C_{3} + C_{2} C_{4} + C_{3} C_{5} + \dots$ - are handled by multiplying two binomial expansions and comparing coefficients of a suitable power of $x$ .

Case I - all terms positive. Multiply

(1 + x)^{n} = C_{0} + C_{1} x + \dots + C_{n} x^{n}

with

(x + 1)^{n} = C_{0} x^{n} + C_{1} x^{n - 1} + \dots + C_{n}

, then compare coefficients of the required power of

x

on both sides of the identity

(1 + x)^{n} (x + 1)^{n} = (1 + x)^{2 n}

.

Case II - alternating signs. Multiply

(1 - x)^{n}

with

(x + 1)^{n}

(or with

(1 + 1/ x)^{n}

appropriately) and compare coefficients.

Solved Example 6

I

is the integer part and

f

the fractional part of

(2 + 3)^{n}

, prove that

(I + f) (1 - f) = 1

and that

I

is an odd integer.

Solution:

Write $(2 + 3)^{n} = I + f$ where $I \in Z$ and $0 \leq f < 1$ .

Note that $(2 + 3) (2 - 3) = 4 - 3 = 1$ , so $(2 + 3)^{n} (2 - 3)^{n} = 1$ . Hence it suffices to prove $(2 - 3)^{n} = 1 - f$ .

Adding the two binomial expansions:

$(2 + 3)^{n} + (2 - 3)^{n} = 2 [2^{n} +^{n} C_{2} 2^{n - 2} \cdot 3 +^{n} C_{4} 2^{n - 4} \cdot 3^{2} + \dots]$

The right side is an even integer because every $3$ term cancels.

Since $0 < 2 - 3 < 1$ , we have $0 < (2 - 3)^{n} < 1$ . Let $f^{'} = (2 - 3)^{n}$ . Then $I + f + f^{'}$ is an even integer, while $0 < f + f^{'} < 2$ . This forces $f + f^{'} = 1$ (the only integer in that range), which gives $f^{'} = 1 - f$ and $I$ = even integer $- 1$ = odd.

Finally, $(I + f) (1 - f) = (2 + 3)^{n} (2 - 3)^{n} = 1$ . $□$

Common Mistakes to Avoid

Watch out

Confusing $\sum C_{r} = 2^{n}$ with $\sum r C_{r} = n \cdot 2^{n - 1}$ - the factor of $r$ changes everything.
Applying Pascal's rule with wrong indices: it is $^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$ , not $^{n + 1} C_{r + 1}$ .
Forgetting that odd-power terms vanish on $\int_{- 1}^{1}$ (used in even-index series proofs).
Overlooking symmetry $C_{k} = C_{n - k}$ when simplifying products of coefficients.
In problems on integer-part/fractional-part, forgetting to check that the conjugate $(2 - 3)^{n}$ lies strictly in $(0, 1)$ .

Frequently Asked Questions

What are the most important properties of binomial coefficients for JEE?

The four most-tested identities for JEE Main and JEE Advanced are: (1) $\sum C_{r} = 2^{n}$ ; (2) $\sum (- 1)^{r} C_{r} = 0$ ; (3) sum of even-indexed = sum of odd-indexed = $2^{n - 1}$ ; (4) Pascal's rule $^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$ . NEET does not include Mathematics, so binomial coefficient properties are relevant only for JEE among the two exams.

How do I sum a series like $C_{1} + 2 C_{2} + 3 C_{3} + \dots + n C_{n}$ ?

Use the differentiation trick: differentiate $(1 + x)^{n} = \sum C_{r} x^{r}$ to get $n (1 + x)^{n - 1} = \sum r C_{r} x^{r - 1}$ , then put $x = 1$ . This yields $C_{1} + 2 C_{2} + \dots + n C_{n} = n \cdot 2^{n - 1}$ . Alternatively use the identity $r \cdot^{n} C_{r} = n \cdot^{n - 1} C_{r - 1}$ .

When should I use integration to sum a binomial coefficient series?

Whenever each term of the series has the form $\frac{C _{r}}{k}$ where $k$ is linear in $r$ (like $\frac{C _{r}}{r + 1}$ or $\frac{C _{r}}{2 r + 1}$ ). Integrate $(1 + x)^{n}$ between suitable limits ( $0$ to $1$ for one-sided series, $- 1$ to $1$ for even-index series). Integration lowers the power of $x$ , matching the $\frac{1}{k}$ factor.

How do I isolate every third coefficient like $C_{0} + C_{3} + C_{6} + \dots$ ?

Use cube roots of unity. If $ω = e^{2 π i /3}$ , then $1 + ω^{k} + ω^{2 k}$ equals $3$ when $k$ is a multiple of $3$ and $0$ otherwise. So $C_{0} + C_{3} + C_{6} + \dots = \frac{( 1 + 1 ) ^{n} + ( 1 + ω ) ^{n} + ( 1 + ω ^{2} ) ^{n}}{3}$ . Similar tricks with $ω$ and $ω^{2}$ multipliers give $C_{1} + C_{4} + \dots$ and $C_{2} + C_{5} + \dots$ .

What is Pascal's rule and why is it useful?

Pascal's rule states $^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$ . It is the recurrence that builds Pascal's triangle row by row. In problems, it lets you telescope sums or rewrite $C_{r - 1} + C_{r}$ as a single coefficient with $(n + 1)$ in place - as in the classic $(C_{0} + C_{1}) (C_{1} + C_{2}) \dots (C_{n - 1} + C_{n})$ product identity.

Why does $C_{0} - C_{1} + C_{2} - \dots + (- 1)^{n} C_{n}$ equal zero?

Substitute $x = - 1$ in the identity $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}$ . The left side becomes $(1 - 1)^{n} = 0$ ; the right side becomes the alternating sum. Hence the alternating sum is exactly $0$ for every $n \geq 1$ .

What is the trick to sum $\sum \frac{C _{r}}{r + 1}$ ?

Use the identity $\frac{^{n} C _{r}}{r + 1} = \frac{^{n + 1} C _{r + 1}}{n + 1}$ . Then $r = 0 \sum n \frac{C _{r}}{r + 1} = \frac{1}{n + 1} r = 0 \sum n^{n + 1} C_{r + 1} = \frac{2 ^{n + 1} - 1}{n + 1}$ . Alternatively integrate $(1 + x)^{n}$ from $0$ to $1$ .

How are products of two binomial coefficients evaluated in JEE Advanced?

The Vandermonde identity approach: write $(1 + x)^{n} (1 + x)^{n} = (1 + x)^{2 n}$ or $(1 + x)^{m} (1 + x)^{n} = (1 + x)^{m + n}$ , expand both sides, and compare coefficients of the required power of $x$ . For example, $\sum_{r = 0}^{n}^{n} C_{r}^{n} C_{r + k} =^{2 n} C_{n - k}$ falls out this way.

Previous year questions on Properties Of Binomial Coefficient

18 questions from past papers, each with a step-by-step solution.

Show all 18 questions

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1. Standard Notation and Basic Properties

2. Series with Alternating Signs

3. Series with a Factor of an Integer (Differentiation Trick)

4. Series with a Coefficient Divided by an Integer (Integration Trick)

5. Series with Product of Two Binomial Coefficients

Common Mistakes to Avoid

Frequently Asked Questions

What are the most important properties of binomial coefficients for JEE?

How do I sum a series like C1​+2C2​+3C3​+⋯+nCn​?

When should I use integration to sum a binomial coefficient series?

How do I isolate every third coefficient like C0​+C3​+C6​+⋯?

What is Pascal's rule and why is it useful?

Why does C0​−C1​+C2​−⋯+(−1)nCn​ equal zero?

What is the trick to sum ∑r+1Cr​​?

How are products of two binomial coefficients evaluated in JEE Advanced?

Previous year questions on Properties Of Binomial Coefficient

Ready to master Binomial Theorem And Mathematical Induction?

How do I sum a series like $C_{1} + 2 C_{2} + 3 C_{3} + \dots + n C_{n}$ ?

How do I isolate every third coefficient like $C_{0} + C_{3} + C_{6} + \dots$ ?

Why does $C_{0} - C_{1} + C_{2} - \dots + (- 1)^{n} C_{n}$ equal zero?

What is the trick to sum $\sum \frac{C _{r}}{r + 1}$ ?