%5E4-formula.jpeg "(a+b+c+d)^4 Formula")
The Formula for (a+b+c+d)^4
When working with algebraic expressions, it's often necessary to expand powers of a binomial or a multinomial. In this article, we'll explore the formula for (a+b+c+d)^4, where a, b, c, and d are variables.
The Binomial Theorem
Before diving into the formula for (a+b+c+d)^4, let's recall the binomial theorem. The binomial theorem states that:
$(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$
where $\binom{n}{k}$ represents the number of combinations of $n$ items taken $k$ at a time.
Multinomial Theorem
The multinomial theorem is an extension of the binomial theorem to more than two variables. It states that:
$(a+b+c+d)^n = \sum_{k_1+k_2+k_3+k_4=n} \binom{n}{k_1, k_2, k_3, k_4} a^{k_1} b^{k_2} c^{k_3} d^{k_4}$
where $\binom{n}{k_1, k_2, k_3, k_4}$ represents the number of combinations of $n$ items taken $k_1, k_2, k_3, k_4$ at a time.
The Formula for (a+b+c+d)^4
Using the multinomial theorem, we can derive the formula for (a+b+c+d)^4 as follows:
$(a+b+c+d)^4 = \sum_{k_1+k_2+k_3+k_4=4} \binom{4}{k_1, k_2, k_3, k_4} a^{k_1} b^{k_2} c^{k_3} d^{k_4}$
There are 15 possible combinations of $k_1, k_2, k_3, k_4$ that add up to 4:
| $k_1$ | $k_2$ | $k_3$ | $k_4$ | $\binom{4}{k_1, k_2, k_3, k_4}$ |
|---|---|---|---|---|
| 4 | 0 | 0 | 0 | 1 |
| 3 | 1 | 0 | 0 | 4 |
| 3 | 0 | 1 | 0 | 4 |
| 3 | 0 | 0 | 1 | 4 |
| 2 | 2 | 0 | 0 | 6 |
| 2 | 1 | 1 | 0 | 12 |
| 2 | 1 | 0 | 1 | 12 |
| 2 | 0 | 2 | 0 | 6 |
| 2 | 0 | 1 | 1 | 12 |
| 2 | 0 | 0 | 2 | 6 |
| 1 | 3 | 0 | 0 | 4 |
| 1 | 2 | 1 | 0 | 12 |
| 1 | 2 | 0 | 1 | 12 |
| 1 | 1 | 2 | 0 | 12 |
| 1 | 1 | 1 | 1 | 24 |
| 1 | 0 | 3 | 0 | 4 |
| 1 | 0 | 2 | 1 | 12 |
| 1 | 0 | 1 | 2 | 12 |
| 1 | 0 | 0 | 3 | 4 |
| 0 | 4 | 0 | 0 | 1 |
| 0 | 3 | 1 | 0 | 4 |
| 0 | 3 | 0 | 1 | 4 |
| 0 | 2 | 2 | 0 | 6 |
| 0 | 2 | 1 | 1 | 12 |
| 0 | 2 | 0 | 2 | 6 |
| 0 | 1 | 3 | 0 | 4 |
| 0 | 1 | 2 | 1 | 12 |
| 0 | 1 | 1 | 2 | 12 |
| 0 | 0 | 4 | 0 | 1 |
| 0 | 0 | 3 | 1 | 4 |
| 0 | 0 | 2 | 2 | 6 |
| 0 |
