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Magazine Chapter 6: Problem 25
Prove $$(a+b+c)^{n}=\sum_{0 \leq i+j \leq n} \frac{n !}{i ! j !(n-i-j) !} a^{i} b^{j} c^{n-i-j}$$.
Short Answer
Expert verified
\[ (a+b+c)^n = \sum_{0 \leq i+j \leq n} \frac{n !}{i ! j !(n-i-j) !} a^{n-i} b^{i-j} c^j \]
Step by step solution
01
Basic Understanding
We know that according to the Binomial theorem, \[ (x+y)^n = \sum_{r=0}^{n} {n \choose r} x^{n-r}y^r \]. Here, we are given three terms inside the bracket instead of two, and the power is 'n', so we can write it as \[ (a+b+c)^n \]. We are required to prove this equals to the given expression.
02
Replacing c with x+y
We can replace 'c' by 'x+y' in the above expression, so that it can be converted into binomial form, i.e., \[ (a+(x+y))^n \]. Now this is in the form of a binomial expression and we can expand it according to the binomial theorem. We therefore get \[ (a+(x+y))^n = \sum_{i=0}^{n} {n \choose i} a^{n-i}(x+y)^i \]
03
Substituting for x and y
In the earlier step, \(x+y\) was in place of 'c'. Now we consider 'x' equal to 'b' and 'y' as 'c' and replace them back, i.e., \( (a+(b+c))^n = \sum_{i=0}^{n} {n \choose i} a^{n-i}(b+c)^i \)
04
Expanding the binomial again
Consider \((b+c)^i\) as a binomial expression and further expand it. Hence, \( (a+(b+c))^n = \sum_{i=0}^{n} {n \choose i} a^{n-i} \sum_{j=0}^{n} {i \choose j} b^{i-j} c^j \)
05
Solving both Sigma
Solving the Sigma expressions results in: \( (a+(b+c))^n = \sum_{i=0}^{j} {n \choose i} {i \choose j} a^{n-i} b^{i-j} c^{j} \). As we have to sum for all i and j such that i+j is less than or equal to n, the equation becomes: \( (a+(b+c))^n = \sum_{0 \leq i+j \leq n} {n \choose i} {i \choose j} a^{n-i} b^{i-j} c^j \)
06
Applying the formula for combination
Rewriting the combination as a factorial for simplification and understanding it as a distribution of 'n' objects in 'r' types, we can then write the equation as: \( (a+b+c)^n = \sum_{0 \leq i+j \leq n} \frac{n !}{i ! j !(n-i-j) !} a^{n-i} b^{i-j} c^j \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Binomial Theorem
The Binomial Theorem provides a formula for expanding expressions raised to a power. It is generally used for expressions containing two terms, like \( (x+y)^n \). This theorem simplifies finding each term in the expansion without having to multiply everything out manually. Each term in the expansion is of the form:
- \( {n \choose r} x^{n-r} y^r \)
- where \( {n \choose r} \) is a binomial coefficient
- that denotes the number of ways to pick \( r \) elements from \( n \) elements
Combinatorics
Combinatorics is the branch of mathematics focusing on counting and arranging objects. It plays a crucial role when working with the Binomial Theorem because it deals with the number of ways to choose elements from a set. In the context of the exercise, it helps in calculating terms involving multiple variables such as \( (a+b+c)^n \). Each specific combination of the indices \(i, j,\text{ and } n-i-j\) corresponds to the number of ways objects could be arranged using these elements:
- The term \( \frac{n!}{i!j!(n-i-j)!} \) comes from combinatorial analysis
- This factorial expression measures how to assign values across different variables
- It reveals the complexity of choosing multiple variables in expansions
Algebraic Expansion
Algebraic expansion involves expanding an expression into a sum of terms to simplify calculations and clarify the outcome. In the problem, the expression \( (a+b+c)^n \) is expanded using the principles of binomial and multinomial expansions. This follows a structured approach:
- Initially, treat \( (a+b+c) \) as \( (a+(b+c)) \)
- Apply the binomial expansion to spread \( (a+(b+c))^n \)
- Further break down \( (b+c)^i \) using the binomial expression
- Each subsequent level involves using combinatorial coefficients
- Ensuring all combinations of terms are considered
