Binomial inside Binomial

Binomial Theorem

The Binomial Theorem gives a direct formula to expand a power of a binomial without performing repeated multiplication. For any positive integer $n$:

$ (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 $

Proof Using Mathematical Induction

Let the statement be:

$ P(n): (a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r $

Base Case

For $n = 1$:

$ P(1): (a + b)^1 = a + b $

So $P(1)$ is true.

Inductive Step

Assume $P(k)$ is true:

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

Now consider $P(k+1)$:

$ (a + b)^{k+1} = (a + b)(a + b)^k $

Expand:

$ (a + b)^{k+1} = \sum_{r=0}^{k} \binom{k}{r} a^{k-r+1} b^0 + \sum_{r=0}^{k} \binom{k}{r} a^{k-r} b^{r+1} $

Group like terms and apply the identity $\binom{k}{r} + \binom{k}{r-1} = \binom{k+1}{r}$:

$ (a + b)^{k+1} = \sum_{r=0}^{k+1} \binom{k+1}{r} a^{k+1-r} b^r $

Thus $P(k+1)$ holds, and by induction the theorem is true for all positive integers $n$.

Binomial Coefficient

A binomial coefficient represents the number of ways to choose $r$ elements from $n$:

$ \binom{n}{r} = \frac{n!}{r!(n-r)!} $

These coefficients appear in every term of $(a+b)^n$ and also in many combinatorial and probability problems.

Binomial Inside Binomial

Nested binomial expressions arise when a binomial appears inside another binomial, such as $(1 + (1+x)^3)^8$ or $(2 + (x+1)^5)^{10}$. To simplify these expressions or evaluate their coefficients, we often need identities involving binomial coefficients, especially those where the upper index varies.

Summation with Changing Upper Index

A common form in binomial-inside-binomial problems is:

$ \sum_{r=0}^{n} \binom{n+r}{r} $

This expands to:

$ \binom{n}{0} + \binom{n+1}{1} + \binom{n+2}{2} + \dots + \binom{2n}{n} $

Step 1: Use symmetry

$ \binom{n+r}{r} = \binom{n+r}{n} $

So the series becomes:

$ \binom{n}{n} + \binom{n+1}{n} + \binom{n+2}{n} + \dots + \binom{2n}{n} $

Step 2: Use Pascal’s identity

Using:

$ \binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r} $

We combine the first two terms:

$ \binom{n}{n} + \binom{n+1}{n} = \binom{n+2}{n+1} $

Repeat this combining process:

$ \binom{n+2}{n+1} + \binom{n+3}{n+1} = \binom{n+3}{n+2} $

Continue until all terms merge into a single term:

$ \sum_{r=0}^{n} \binom{n+r}{r} = \binom{2n+1}{n+1} $

This is a very important identity for simplifying multi-layered binomial expressions.

Why Binomial Inside Binomial Matters

This identity is frequently used in:

• coefficient extraction when binomials are nested

• simplifying long combinatorial sums

• solving Olympiad and JEE problems

• expressions where powers of a binomial appear inside another power

Nested expansions like $(1 + (1+x)^n)^m$ often reduce to sums of the form $\sum \binom{n+r}{r}$.

Example of How It Appears in Problems

Consider the nested binomial:

$ (1 + (1+x)^n )^n $

To find the coefficient of $x^k$, we expand twice. This produces sums involving binomial coefficients whose upper index changes with $r$. Those sums simplify using the identity:

$ \sum_{r=0}^{n} \binom{n+r}{r} = \binom{2n+1}{n+1} $

This drastically reduces the computation.

Useful Properties for Binomial Inside Binomial Problems

We have given below the important and useful properties related to binomial inside binomial, so that you can make your coplex calculations easier and faster.

Symmetry Property

$ \binom{n}{r} = \binom{n}{n-r} $

Helps rewrite sums in a more convenient form.

Pascal Identity

$ \binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r} $

This is the key resource used for collapsing long summations.

Binomial Theorem for Any Index

Used when the inner binomial is fractional or negative.

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

Important Results

• Largest coefficient estimation

• Term location using ratios

• Remainder and last-digit evaluation from binomial expansion

These concepts help handle competitive-exam-level nested expressions.

Solved Examples Based on Binomial inside Binomial

Example 1:
If $(1+x)^n = C_0 + C_1 x + \ldots + C_n x^n$, then

$C_0 \cdot {}^{2n}C_n - C_1 \cdot {}^{2n-1}C_n + C_2 \cdot {}^{2n-2}C_n - C_3 \cdot {}^{2n-3}C_n + \ldots + (-1)^n C_n \cdot {}^n C_n$ is equal to:

1. $\frac{1}{n}$

2. $2^{-n}$

3. $2^n$

4. $1$

Solution:

$C_0 \cdot {}^{2n}C_n - C_1 \cdot {}^{2n-1}C_n + C_2 \cdot {}^{2n-2}C_n - C_3 \cdot {}^{2n-3}C_n + \ldots + (-1)^n C_n \cdot {}^n C_n$

$=\text{ coefficient of } x^n \text{ in }$

$\left[C_0(1+x)^{2n} - C_1(1+x)^{2n-1} + C_2(1+x)^{2n-2} - \ldots + (-1)^n {}^nC_n (1+x)^n \right]$

Coefficient of $x^n$ in:

$(1+x)^n \left[ C_0(1+x)^n - C_1(1+x)^{n-1} + C_2(1+x)^{n-2} - \ldots + (-1)^n {}^nC_n \cdot 1 \right]$

$=\text{ coefficient of } x^n \text{ in } (1+x)^n \left[( (1+x)-1 )^n \right]$

$=\text{ coefficient of } x^n \text{ in } (1+x)^n \cdot x^n$

$= \text{ constant term in } (1+x)^n = 1$

Hence, the answer is option 4.

Example 2:
In a hockey series between teams $X$ and $Y$, they play until one team wins $m$ matches. The number of ways in which team $X$ wins is:

1. $2^m$

2. ${}^{2m}P_m$

3. ${}^{2m}C_m$

4. None of these

Solution:

Team $X$ wins if it wins the $(m+r)$-th match and wins $(m-1)$ matches from the first $(m+r-1)$ matches.

Total number of ways:

$\sum_{r=0}^{m} {}^{m+r-1}C_{m-1} = \frac{{}^{2m}C_m}{2}$

Hence, the answer is option 4.

Example 3:
The sum $S_n = \sum_{k=0}^{n} (-1)^k \cdot {}^{3n}C_k$, for $n = 1,2,\ldots$, is:

1. $(-1)^n \cdot {}^{3n-1}C_{n-1}$

2. $(-1)^n \cdot {}^{3n-1}C_n$

3. $(-1)^n \cdot {}^{3n-1}C_{n+1}$

4. None

Solution:

$S_n = {}^{3n}C_0 - {}^{3n}C_1 + {}^{3n}C_2 - {}^{3n}C_3 + \ldots + (-1)^n {}^{3n}C_n$

Now apply:

${}^{3n}C_k = {}^{3n-1}C_{k} + {}^{3n-1}C_{k-1}$ with sign adjustments.

After pairing and cancellation:

$S_n = (-1)^n \cdot {}^{3n-1}C_n$

Hence, the answer is option 2.

Example 4:
The value of
${}^{50}C_4 + \sum_{r=1}^{6} {}^{56-r}C_3$ is:

1. ${}^{55}C_3$

2. ${}^{55}C_4$

3. ${}^{56}C_4$

4. ${}^{56}C_3$

Solution:

${}^{50}C_4 + \left( {}^{50}C_3 + {}^{51}C_3 + {}^{52}C_3 + \ldots + {}^{55}C_3 \right)$

Combine terms using:

${}^{n}C_r + {}^{n}C_{r-1} = {}^{n+1}C_r$

Repeatedly applying this property collapses the sequence into:

${}^{56}C_4$

Hence, the answer is option 3.

Example 5:
Evaluate:
${}^{n+1}C_2 + 2\left[ {}^{2}C_2 + {}^{3}C_2 + {}^{4}C_2 + \ldots + {}^{n}C_2 \right]$

1. $\sum n^3$

2. $\sum n^2$

3. $\sum n$

4. $\frac{n+1}{2}$

Solution:

Expression:

$={}^{n+1}C_2 + 2\left[ {}^{2}C_2 + {}^{3}C_2 + \ldots \right]$

Rewrite each group using:

${}^{k}C_2 = {}^{k-1}C_1 + {}^{k-1}C_2$

Eventually this reduces to:

$={}^{n+1}C_2 + 2,{}^{n+1}C_3$

$={}^{n+1}C_2 + {}^{n+1}C_3 + {}^{n+1}C_3$

$={}^{n+2}C_3 + {}^{n+1}C_3$

$=\frac{(n+2)(n+1)n}{6} + \frac{(n+1)n(n-1)}{6}$

$=\frac{n(n+1)(2n+1)}{6}$

$=\sum n^2$

Hence, the answer is option 2.

Last Digits and Remainder using the Binomial Expansion