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Binomial inside Binomial

Binomial inside Binomial

Edited By Komal Miglani | Updated on Jul 02, 2025 08:02 PM IST

An expression with two terms is called the binomial expansion. In the case of higher degree expression, it is difficult to calculate it manually. In these cases, Binomial theorem can be used to calculate it. Binomial theorem is used for the expansion of a binomial expression with a higher degree. Binomial coefficients are the coefficients of the terms in the Binomial expansion. Binomial theorem is proved using the concept of mathematical induction. Apart from Mathematics, Binomial theorem is also used in statistical and financial data analysis.

Binomial inside Binomial
Binomial inside Binomial

This article is about the binomial inside binomial which falls under the broader category of Binomial Theorem and its applications. It is one of the important topics for competitive exams.

Binomial Theorem

Statement:

If $n$ is any positive integer, then

$ (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 - 1} a b^{n - 1} + \binom{n}{n} b^n $

Proof:

The proof is obtained by applying the principle of mathematical induction.

Let the given statement be:

$ P(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-1} a b^{n-1} + \binom{n}{n} b^n $

For $ n = 1 $, we have:

$ P(1): (a + b)^1 = \binom{1}{0} a^1 + \binom{1}{1} b^1 = a + b $

Thus, $ P(1) $ is true.

Suppose $ P(k) $ is true for some positive integer $ k $, i.e.,

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

We shall prove that $ P(k + 1) $ is also true, i.e.,

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

Now,

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

$ = (a + b) \left[\binom{k}{0} a^k + \binom{k}{1} a^{k-1} b + \binom{k}{2} a^{k-2} b^2 + \dots + \binom{k}{k-1} a b^{k-1} + \binom{k}{k} b^k\right] $

[from (1)]

$ = \binom{k}{0} a^{k+1} + \binom{k}{1} a^k b + \binom{k}{2} a^{k-1} b^2 + \dots + \binom{k}{k-1} a^2 b^{k-1} + \binom{k}{k} a b^k $

$ + \binom{k}{0} a^k b + \binom{k}{1} a^{k-1} b^2 + \binom{k}{2} a^{k-2} b^3 + \dots + \binom{k}{k-1} a b^k + \binom{k}{k} b^{k+1} $

[by actual multiplication]

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

[grouping like terms]

$ = \binom{k+1}{0} a^{k+1} + \binom{k+1}{1} a^k b + \binom{k+1}{2} a^{k-1} b^2 + \dots + \binom{k+1}{k} a b^k + \binom{k+1}{k+1} b^{k+1}$

(by using $ \binom{k+1}{0} = 1 $, $ \binom{k}{r} + \binom{k}{r-1} = \binom{k+1}{r} $, and $ \binom{k}{k} = 1 = \binom{k+1}{k+1} $)

Thus, it has been proved that $ P(k + 1) $ is true whenever $ P(k) $ is true. Therefore, by the principle of mathematical induction, $ P(n) $ is true for every positive integer $ n $.


Binomial Coefficient

The combination $\binom{n}{r}$ or ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}$ occuring in the Binomial theorem is called a Binomial coefficient, where $\binom{n}{r}=C(n, r)={ }^n C_r=\frac{n!}{r!(n-r)!}$.

Binomial inside Binomial

Let's see some examples to understand about combining binomials.

How do we find the summation of the series where upper index also varies e.g., $\sum_{r=0}^n{ }^{n+r} C_r={ }^n C_0+{ }^{n+1} C_1+{ }^{n+2} C_2+\cdots+{ }^{n+n} C_n$

[put value of $r=0,1,2, \ldots \ldots$, upto n]

By using the property of the binomial coefficient

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

$\sum_{r^n=0}^{\mathrm{n}+{ }^{\mathrm{r}}} \mathrm{C}_{\mathrm{r}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}+{ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{n}}+{ }^{\mathrm{n}+2} \mathrm{C}_{\mathrm{n}}+\cdots+{ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{n}} $

$\text { Now as }{ }^{\mathrm{n}} C_n={ }^{n+1} C_{n+1}=1 \text {, so we can write } $

$\sum_{\mathrm{r}=0}^{\mathrm{n}}{ }^{\mathrm{n}+\mathrm{r}} \mathrm{C}_{\mathrm{r}}={ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{n}+1}+{ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{n}}+{ }^{\mathrm{n}+2} \mathrm{C}_{\mathrm{n}}+\cdots+{ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{n}} $

Now first $2$ terms can be combined by using the property

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

$\sum_{r=0}^{\mathrm{n}}{ }^{\mathrm{n}+\mathrm{r}} C_r={ }^{\mathrm{n}+2} C_{\mathrm{n}+1}+{ }^{\mathrm{n}+2} C_n+\cdots+{ }^{2 n} C_n $

now first two terms can again be combined using the same property, and this process can be continued till last term is also combined

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

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Solved Examples Based on Binomial inside Binomial

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

$C_0{ }^{2 n} C_n-C_1 \cdot{ }^{2 n-1} C_n+C_2 \cdot{ }^{2 n-2} C_n-C_3 \cdot{ }^{2 n-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:

Binomial Inside Binomial

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

$=\text { Coefficient of }{\underset{\sim}{x}}^n \text { in } $

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

$\left.\quad-C_3(1+x)^{2 n-3}+\ldots+(-1)^n{ }^n C_n \cdot(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}\right. $

$\left.-C_3(1+x)^{n-3}+\ldots+(-1)^n \cdot{ }^n C_n \cdot 1\right] $

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

$=\text { Coefficient of }{\underset{\sim}{\mathrm{X}}}_{\mathrm{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 decide to play till a team wins ' $m$ ' matches. Then the number of ways in which team $X$ wins is

1) $2^m$

2) ${ }^{2 m} P_m$

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

4) None of these

Solution:

Theorem of Combination-

Each of the different groups or selections which can be made by taking r things from $n$ things is called a combination.

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

wherein

Where $1 \leq r \leq n$

Team X will win if it wins $(m+r)$ th match and

wins $m-1$ match from the first $m+r-1$ matches,

Thus, the total number of ways $=\sum_{r=0}^m{ }^{m+r-1} C_{m-1}=\frac{{ }^{2 m} C_m}{2}$

Hence, the answer is option 4.

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

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

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

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

4) None of these

Solution:

$\mathrm{S}_{\mathrm{n}}={ }^{3 n} \mathrm{C}_0-{ }^{3 n} \mathrm{C}_1+{ }^{3 n} \mathrm{C}_2+\ldots \ldots+(-1)^n \cdot{ }^{3 n} \mathrm{C}_{\mathrm{n}} $

$\text { But }{ }^{3 n} \mathrm{C}_0={ }^{3 n-1} \mathrm{C}_0 $

$-{ }^{3 n} \mathrm{C}_1=--{ }^{3 n-1} \mathrm{C}_0-{ }^{3 n-1} \mathrm{C}_1 $

${ }^{3 n} \mathrm{C}_2={ }^{3 n-1} \mathrm{C}_1+{ }^{3 n-1} \mathrm{C}_2 $

$-{ }^{3 n} \mathrm{C}_3=--{ }^{3 n-1} C_2-{ }^{3 n-1} \mathrm{C}_3 $

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

On adding we get

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

1) ${ }^{55} \mathrm{C}_3$

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 \ldots+{ }^{55} C_3\right) $

Taking the first two terms together and adding them and following the same pattern, we get ${ }^{56} C_4$.

$\left[\mathrm{As}^n C_r+{ }^n C_{r-1}={ }^{n+1} C_r\right] $

Hence, the answer is option (3).

Example 5: The value of the expression ${ }^{(n+1)} C_2+2\left[{ }^2 C_2+{ }^3 C_2+{ }^4 C_2+\ldots . .+{ }^n C_2\right]$ is:

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+{ }^4 C_2+\ldots \ldots .\right] $

$={ }^{(n+1)} C_2+2\left[{ }^4 C_3+{ }^4 C_2+\ldots \ldots . .\right] $

$={ }^{(n+1)} C_2+2\left[{ }^5 C_3+{ }^5 C_2+\ldots \ldots .+{ }^n C_2\right] $

$={ }^{(n+1)} C_2+2{ }^{(n+1)} C_3 \text { ultimately } $

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

Hence, the answer is option (2).



Frequently Asked Questions (FAQs)

1. What is a "binomial inside a binomial" in the context of the Binomial Theorem?
A "binomial inside a binomial" refers to a situation where one or both terms of a binomial expression contain another binomial. For example, in the expression (a + (b + c))^n, we have a binomial (b + c) inside the main binomial. This creates a more complex structure that requires careful application of the Binomial Theorem.
2. How does the presence of a binomial inside another binomial affect the expansion process?
When a binomial is inside another binomial, the expansion process becomes more intricate. You need to treat the inner binomial as a single term and expand the outer binomial first. Then, you may need to further expand terms containing the inner binomial. This often results in more terms and a more complex final expression.
3. Can the Binomial Theorem be directly applied to expressions with binomials inside binomials?
The Binomial Theorem cannot be directly applied to the entire expression when there's a binomial inside a binomial. Instead, you need to apply it in stages: first to the outer binomial, treating the inner binomial as a single term, and then potentially to terms containing the inner binomial if further expansion is needed.
4. What is the first step in expanding (a + (b + c))^n?
The first step is to treat (b + c) as a single term and apply the Binomial Theorem to (a + (b + c))^n. This gives you terms of the form a^k((b + c)^(n-k)), where k ranges from 0 to n.
5. How do you handle terms like (b + c)^m that result from the initial expansion of a binomial inside a binomial?
Terms like (b + c)^m can be further expanded using the Binomial Theorem if m is a positive integer. If m is large, you might choose to leave it in this compact form or expand it depending on the requirements of the problem.
6. What is the key difference between expanding (a + b)^n and (a + (b + c))^n?
The key difference is that (a + b)^n results in a sum of n+1 terms, while (a + (b + c))^n initially results in n+1 terms, but some of these terms contain (b + c) raised to various powers, which may require further expansion.
7. How does the complexity of the expansion change as n increases in (a + (b + c))^n?
As n increases, the complexity of the expansion grows significantly. Not only do you have more terms from the initial expansion, but each term containing (b + c) potentially needs to be expanded further, leading to a rapid increase in the number of terms in the final expression.
8. What is the significance of the binomial coefficients in expanding expressions with binomials inside binomials?
Binomial coefficients play a crucial role in determining the number of ways to choose terms from both the outer and inner binomials. They appear in both the initial expansion and any subsequent expansions of inner binomial terms.
9. How would you approach expanding (a + (b - c))^3?
To expand (a + (b - c))^3:
10. What is the Pascal's Triangle and how is it related to binomials inside binomials?
Pascal's Triangle is a triangular array of binomial coefficients. While it's directly applicable to simple binomial expansions, for binomials inside binomials, you may need to use Pascal's Triangle multiple times: once for the outer binomial expansion and then for each inner binomial term that needs further expansion.
11. How does the distributive property come into play when dealing with binomials inside binomials?
The distributive property is crucial when expanding binomials inside binomials. After the initial expansion of the outer binomial, you often need to distribute powers to the terms of the inner binomial. For example, in expanding (a + (b + c))^2, you'll need to distribute the square to (b + c) in the term (b + c)^2.
12. What is the general form of the expansion of (a + (b + c))^n?
The general form is:
13. How do you determine when to stop expanding in a binomial inside binomial problem?
The decision to stop expanding depends on the problem requirements and the level of detail needed. Generally, you expand until you have terms with single variables or constants. However, in some cases, leaving some terms in a compact form like (b + c)^k might be more useful or required.
14. What common mistakes do students make when expanding binomials inside binomials?
Common mistakes include:
15. How does the presence of coefficients in the inner binomial, like (a + (2b - 3c))^n, affect the expansion process?
Coefficients in the inner binomial don't change the overall process, but they do affect the final terms. You still expand the outer binomial first, treating (2b - 3c) as a single term. When expanding terms involving the inner binomial, these coefficients will be raised to powers and multiplied accordingly.
16. Can you use the Binomial Theorem for expressions like (a + b + c)^n?
The Binomial Theorem is specifically for expressions with two terms. For (a + b + c)^n, you would need to use the Multinomial Theorem, which is a generalization of the Binomial Theorem for expressions with more than two terms.
17. How does expanding (a - (b + c))^n differ from expanding (a + (b + c))^n?
The main difference is in the signs of the terms. In (a - (b + c))^n, the odd powers of (b + c) will have a negative sign. The expansion would look like:
18. What is the relationship between the terms in the expansion of (a + (b + c))^n and (a + b + c)^n?
While both expansions will contain similar terms, (a + (b + c))^n will group b and c together in its initial expansion, potentially leaving some terms unexpanded. (a + b + c)^n, expanded using the Multinomial Theorem, will fully separate a, b, and c from the start, resulting in a different grouping of terms.
19. How do you verify the correctness of your expansion of a binomial inside a binomial?
To verify your expansion:
20. What is the effect of having variables with exponents in the inner binomial, like (x + (y^2 + z^3))^n?
Variables with exponents in the inner binomial don't change the expansion process, but they do affect the final terms. When you expand terms containing (y^2 + z^3), these exponents will be further multiplied. For example, (y^2 + z^3)^2 would include terms like y^4, y^2z^3, and z^6.
21. How does the concept of binomials inside binomials extend to more complex nested structures, like (a + (b + (c + d)))^n?
For more complex nested structures, you apply the expansion process recursively, starting from the outermost binomial and working inwards. Each level of nesting adds a layer of complexity and potentially increases the number of terms in the final expansion significantly.
22. Can you apply the binomial series expansion to expressions with binomials inside binomials when n is not a positive integer?
Yes, the binomial series can be applied when n is not a positive integer, but it results in an infinite series rather than a finite expansion. However, this becomes much more complex with binomials inside binomials and is generally handled using advanced calculus techniques.
23. How do you simplify the expansion of (1 + (x + y))^n for large values of n?
For large n, fully expanding (1 + (x + y))^n can be impractical. Instead, you might:
24. What is the role of the Binomial Theorem in probability problems involving nested events?
In probability, binomials inside binomials can represent nested events. The Binomial Theorem helps calculate probabilities of combined outcomes. For example, (p + (1-p))^n might represent n trials where each trial has a sub-structure of possible outcomes.
25. How does factoring play a role in simplifying expansions of binomials inside binomials?
Factoring can help simplify complex expansions by identifying common terms. After expanding, look for common factors among terms. Sometimes, factoring can reveal a simpler structure or a recognizable pattern in the expansion.
26. What strategies can be used to mentally estimate the result of expanding a binomial inside a binomial without full calculation?
To mentally estimate:
27. How does the expansion of (a + (b + c))^n relate to combinatorial problems?
The expansion relates to combinatorics in several ways:
28. What is the significance of the term (b + c)^n in the expansion of (a + (b + c))^n?
The term (b + c)^n is significant because:
29. How does the presence of a binomial inside a binomial affect the growth rate of terms in the expansion as n increases?
The growth rate of terms increases more rapidly than in a simple binomial expansion. As n increases:
30. What is the connection between binomials inside binomials and nested summations in mathematics?
Binomials inside binomials often lead to nested summations in their expanded form. The outer summation comes from expanding the outer binomial, while inner summations arise from expanding terms containing the inner binomial. This nested structure mirrors the nested nature of the original expression.
31. How would you approach teaching the concept of binomials inside binomials to students who have just learned the basic Binomial Theorem?
To teach this concept:
32. What real-world applications involve calculations with binomials inside binomials?
Real-world applications include:
33. How does the concept of binomials inside binomials extend to polynomials inside binomials, like (x + (ay^2 + by + c))^n?
The approach is similar:
34. What is the effect of having a constant term in the inner binomial, like (x + (y + 1))^n?
A constant in the inner binomial doesn't change the expansion process but affects the final terms:
35. How do you determine the degree of the resulting polynomial when expanding a binomial inside a binomial?
To determine the degree:
36. What is the relationship between the coefficients in the expansion of (a + (b + c))^n and those in (a + b + c)^n?
While both expansions will have some coefficients in common:
37. How does the presence of a binomial inside a binomial affect the symmetry of the expansion compared to a simple binomial expansion?
The presence of an inner binomial often reduces the symmetry:

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