Sum of Binomial Coefficients

Binomial Expression: Definition, Example

A binomial expression is simply an algebraic expression that contains exactly two terms. You see them all the time in algebra — expressions like
$ (a+b)^2 $,
$ \left(\sqrt{x}+\frac{k}{x^2}\right)^5 $,
$ (x+9y)^{-2/3} $.

Even though these expressions look small, raising them to higher powers can quickly become complicated. Imagine trying to expand $(x+y)^{52}$ by multiplying $(x+y)$ again and again — nobody has that kind of patience.

Understanding Patterns Through Simple Binomial Expansions

Before jumping to large powers, it helps to observe how smaller powers behave:

$
(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
$

If you look closely:

• The exponent of $x$ decreases by 1 in each term.

• The exponent of $y$ increases by 1 in each term.

• The sum of exponents in each term is always n.

• The coefficients follow a predictable pattern:
  $ \binom{n}{0}, \binom{n}{1}, \binom{n}{2}, \dots, \binom{n}{n} $

These exactly match the numbers found in Pascal’s Triangle, and they form the heart of the Binomial Theorem.

The Binomial Theorem

When $n$ is a positive integer, the expansion of $(a+b)^n$ can be written directly as:

$ (a+b)^n = \binom{n}{0} a^n + \binom{n}{1} a^{n-1}b + \binom{n}{2} a^{n-2} b^2 + \dots + \binom{n}{n} b^n $

This formula saves you from endless multiplication and is foundational in algebra, combinatorics, and data science applications.

What Is a Binomial Coefficient?

The coefficient
$ \binom{n}{r} = \frac{n!}{r!(n-r)!} $
is called a binomial coefficient.

It represents the number of ways to choose r objects out of n, making it extremely useful in probability, counting problems, statistics, and even machine learning model calculations.

Key Results on the Sum of Binomial Coefficients

These identities help simplify expressions and solve higher-level problems in competitive exams like JEE, CUET, and in real-world computation.

1. Sum of All Binomial Coefficients

A beautiful identity states:

$ \sum_{r=0}^n {^nC_r} = 2^n $

This comes from simply substituting $a = 1$ and $b = 1$ in the Binomial Theorem:

$(1+1)^n = 2^n$

This result is widely used in:

• probability distributions

• algorithmic complexity

• combinatorial counting

2. Alternating Binomial Sum

Surprisingly, the alternating sum equals zero:

$ \sum_{r=0}^n (-1)^r {^nC_r} = 0 $

This happens because terms pair up as $T_k$ and $T_{n-k}$, canceling each other out.

You’ll see this identity used in:

• evaluating alternating series

• solving recursive relations

• simplifying probability expressions

3. Even vs Odd Binomial Coefficients

Both the sums of even-indexed and odd-indexed binomial coefficients equal:

$ 2^{n-1} $

Meaning:

• $C_0 + C_2 + C_4 + \dots = 2^{n-1}$

• $C_1 + C_3 + C_5 + \dots = 2^{n-1}$

This symmetry is extremely helpful when working with balanced outcomes, such as in:

• coin-tossing experiments

• symmetric probability distributions

• partitioning data into equal-weight categories

Key Formulae of Binomial Theorem

Provided below are the important formulae related to the binomial theorem:

Why These Binomial Identities Matter in Real Life

These formulas aren’t just abstract math — they help solve real-life complex problems such as:

• calculating the number of possible datasets in machine learning models

• analyzing network combinations in computer science

• computing probabilities in finance and risk analysis

• optimizing resource allocation problems

• designing encryption and secure password systems

Whenever you need to count possibilities or understand how quantities grow, binomial coefficients and their sums step in as powerful resources.

Solved Examples Based on Sum of Binomial Coefficients

Example 1:
Let
$S_n = 1 + q + q^2 + \ldots + q^n$
and
$T_n = 1 + \left(\frac{q+1}{2}\right) + \left(\frac{q+1}{2}\right)^2 + \ldots + \left(\frac{q+1}{2}\right)^n$
where $q$ is a real number and $q \neq 1$.

If
${ }^{101} C_1 + { }^{101} C_2 S_1 + \ldots + { }^{101} C_{101} S_{100} = \alpha T_{100}$
then $\alpha$ is equal to:

1. 200

2. $2^{100}$

3. 2021

4. $2^{99}$

Solution
Sum of binomial coefficients:
$C_0 + C_1 + C_2 + \ldots + C_n = 2^n$

$T_n = 1 + \left(\frac{q+1}{2}\right) + \left(\frac{q+1}{2}\right)^2 + \cdots + \left(\frac{q+1}{2}\right)^n$

$T_n = \frac{\left(\frac{q+1}{2}\right)^{n+1} - 1}{\left(\frac{q+1}{2}\right)-1}$

Given:
${ }^{101}C_1 + { }^{101}C_2 S_1 + \cdots + { }^{101}C_{101} S_{100} = \alpha T_{100}$

$\alpha T_{100} = \sum_{r=1}^{101} { }^{101}C_r \cdot S_{r-1}$

$\alpha T_{100} = \sum_{r=1}^{101} { }^{101}C_r \left[\frac{q^r - 1}{q - 1}\right]$

$\alpha T_{100} = \frac{1}{q-1} \left(\sum_{r=1}^{101} { }^{101}C_r (q^r - 1)\right)$

$\alpha T_{100} = \frac{1}{q-1} \left((1+q)^{101} - 1 - (2^{101} - 1)\right)$

$\alpha T_{100} = \frac{(1+q)^{101} - 2^{101}}{q - 1}$

$\frac{\alpha \left(\left(\frac{q+1}{2}\right)^{101} - 1\right)}{\left(\frac{q+1}{2}\right)-1} = \frac{(1+q)^{101} - 2^{101}}{q-1}$

$\frac{2}{2^{101}} \alpha \left(\frac{(q+1)^{101} - 2^{101}}{(q+1)-2}\right) = \frac{(1+q)^{101} - 2^{101}}{q-1}$

$\alpha \left(\frac{1}{2^{100}}\right) = 1$

$\alpha = 2^{100}$

Correct option: 2.

Example 2:
If the number of terms in the expansion of
$\left(1 - \frac{2}{x} + \frac{4}{y}\right)^n$
(where $x \neq 0, y \neq 0$) is $28$, then the sum of coefficients is:

1. $64$

2. $2187$

3. $243$

4. $729$

Solution

Using multinomial theorem:
${ }^{n+2} C_2 = 28$

$\frac{(n+2)!}{2! , n!} = 28$

$\frac{(n+2)(n+1)n!}{2n!} = 28$

$(n+2)(n+1) = 56$

Thus $n = 6$.

Sum of coefficients found by putting $x=1, y=1$:

$(1 - 2 + 4)^6 = 3^6 = 729$

Correct option: 4.

Example 3:
If the fractional part of
$\frac{2^{403}}{15}$
is $\frac{k}{15}$, then $k$ equals:

1. 6

2. 8

3. 4

4. 14

Solution

$2^{403} = 2^3 \cdot 2^{400} = 8(2^4)^{100} = 8(15+1)^{100}$

$\frac{8}{15}(15+1)^{100} = \frac{8}{15}(15\lambda + 1)$

$= 8\lambda + \frac{8}{15}$

Fractional part = $\frac{8}{15}$
So $k = 8$.

Correct option: 2.

Example 4:
What is the sum of the coefficients of
$(\sqrt{2} - \sqrt[3]{3} + \sqrt[6]{5})^{10}$?

1. $3^{10}$

2. $2^{10}$

3. 1

4. None

Solution

For $(x+y+z)^n$, sum of coefficients = $(1+1+1)^n = 3^n$.

To determine the coefficient in question, substitute:
$x_1 = 1, x_2 = -1, x_3 = 1, n = 10$

$(1 - 1 + 1)^{10} = 1^{10} = 1$

Correct option: 3.

Example 5:
If $C_r = { }^{25}C_r$ and
$C_0 + 5C_1 + 9C_2 + \cdots + 101 C_{25} = 2^{25} k$,
then $k$ equals:

1. 15

2. 30

3. 51

4. 27

Solution

General term:
$(4r + 1)\cdot { }^{25}C_r$

Applying summation:

$= 4 \sum_{r=0}^{25} r \cdot { }^{25}C_r + 2^{25}$

$= 4 \cdot 25 \cdot 2^{24} + 2^{25}$

$= 50 \cdot 2^{25} + 2^{25}$

$= 51 \cdot 2^{25}$

Thus $k = 51$.

Correct option: 3.

Last Digits and Remainder using the Binomial Expansion