| n | \( a^n - b^n \) |
| 2 | \( (a - b)(a + b) \) |
| 3 | \( (a - b)(a^2 + ab + b^2) \) |
| 4 | \( (a^2 - b^2)(a^2 + b^2) \) |
| 5 | \( (a - b)(a^4 + a^3 b + a^2 b^2 + ab^3 + b^4) \) |
| n | \( a^n + b^n \) |
| 2 | \( (a + b)^2 - 2ab \) |
| 3 | \( (a + b)(a^2 - ab + b^2) \) |
| 4 | \( (a^2 + 2ab + b^2)^2 - 4a^2 b^2 \) |
| n | \( (a-b)^n \) |
| 2 | \(a^2 - 2ab + b^2\) |
| 3 | \(a^3 - 3a^2b + 3ab^2 - b^3\) |
| 4 | \(a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4\) |
| 5 | \(a^5 - 5a^4b + 10a^3b^2 - 10a^2b^3 + 5ab^4 - b^5\) |
| n | \( (a+b)^n \) |
| 2 | \(a^2 + 2ab + b^2\) |
| 3 | \(a^3 + 3a^2b + 3ab^2 + b^3\) |
| 4 | \(a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4\) |
| 5 | \(a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5\) |
| Expression | Expansion |
| \((a + b + c)^2\) | \( \small a^2 + b^2 + c^2 + 2ab + 2ac + 2bc \) |
| \((a - b - c)^2\) | \( \small a^2 + b^2 + c^2 - 2ab - 2ac + 2bc\) |
| \((x + a)(x + b)\) | \(x^2 + (a + b)x + ab\) |
| \((x - a)(x - b)\) | \(x^2 - (a + b)x + ab\) |
$$ (x + a)^n = \sum_{k=0}^{n} \binom{n}{k} x^k a^{n-k} $$ binomial theorem, which states that for any non-negative integer \(n\), the expansion of \( (x+a)^n \) is given by the sum of the terms involving \(x\) and \(a\), each raised to some power:
Here: \(n\) müsbət tam ədəd.
\(n\) is a non-negative integer.
\( \sum \) represents the summation symbol, which means to sum the terms generated by the formula inside the parentheses for \(k = 0\) to \(n\).
\( \binom{n}{k}\) or \( _nC_k\) (also written as C(n,k) or "n choose k") represents the binomial coefficient, which is the number of ways to choose \(k\) items from a set of \(n\) items. It is calculated using the formula: $$ \binom{n}{k} = \frac{n!}{k!(n-k)!} $$ where "\(!\)" denotes the factorial function.
\( x^k \) and \(a^{n-k} \) are the terms involving \(x\) and \(a\), each raised to some power.
In the expansion of \((x+a)^n\), there are \((n+1)\) terms, with each term being a product of the binomial coefficient, \(x\) raised to some power, and \(a\) raised to some other power. The powers of \(x\) and \(a\) decrease and increase, respectively, as we move from the first term to the last term in the expansion.
For example, \((x+a)^3 \): $$ (x+a)^3 = (nC0) x^3 a^0+ (nC1) x^2 a^1+ (nC2) x^1 a^2+ (nC3) x^0 a^3 $$ Using the binomial coefficients: $$ (x+a)^3 = 1(x^3 )(a^0)+3(x^2 )(a^1 )+3(x^1 )(a^2 )+ 1(x^0 )(a^3 )= x^3+3x^2 a+3xa^2+a^3 $$
This formula allows us to easily expand binomials raised to any power without having to manually apply the distributive property multiple times.
