Question

(а+b)^8 вывести формулы

Answer 1

(a+b)^8 = (a+b)(a+b)(a+b)(a+b)(a+b)(a+b)(a+b)(a+b) = ( a² + 2ab + b²) ( a² + 2ab + b²) ( a² + 2ab + b²) ( a² + 2ab + b²) = ( a⁴ + 2a³b + a²b² + 2a³b + 4a²b² + 2ab³ + b²a² + 2ab³ +b⁴ )( a⁴ + 2a³b + a²b² + 2a³b + 4a²b² + 2ab³ + b²a² + 2ab³ +b⁴ ) = (a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴)(a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴) = a^8 + 4a^7×b + 6a^6×b² + 4a^5×b³ + a⁴b⁴ + 4a^7×b + 16a^6×b² + 24a^5×b³ + 16a⁴b⁴ + 4a³b^5 + 6a^6×b² + 24a^5×b³ + 36a⁴b⁴ + 24a³b^5 + 6a²b^6 + 4a^5×b³ + 16a⁴b⁴ + 24a³b^5 + 16a²b^6 + 4ab^7 + a⁴b⁴ + 4a³b^5 + 6a²b^6 + 4ab^7 + b^8 = a^8 + 8a^7×b + 28a^6×b² + 56a^5×b³ + 70a⁴b⁴ + 56a³b^5 + 28a²b^6 + 8ab^7 + b^8

Answer 2

To expand the binomial expression (a+b)^8, you can use the binomial theorem, which states that:

(a + b)^n = Σ (n choose k) * a^(n-k) * b^k

where:

• n is the exponent (in this case, 8)
• k is the index of summation, ranging from 0 to n
• (n choose k) is the binomial coefficient, which is given by (n! / (k! * (n-k)!))
• a and b are the variables (in this case, a and b)

Now, let's apply the binomial theorem to (a + b)^8:

(a + b)^8 = Σ (8 choose k) * a^(8-k) * b^k

where k ranges from 0 to 8.

Let's calculate the terms for k = 0 to k = 8:

1. For k = 0: (8 choose 0) * a^(8-0) * b^0 = 1 * a^8 * 1 = a^8

2. For k = 1: (8 choose 1) * a^(8-1) * b^1 = 8 * a^7 * b

3. For k = 2: (8 choose 2) * a^(8-2) * b^2 = 28 * a^6 * b^2

4. For k = 3: (8 choose 3) * a^(8-3) * b^3 = 56 * a^5 * b^3

5. For k = 4: (8 choose 4) * a^(8-4) * b^4 = 70 * a^4 * b^4

6. For k = 5: (8 choose 5) * a^(8-5) * b^5 = 56 * a^3 * b^5

7. For k = 6: (8 choose 6) * a^(8-6) * b^6 = 28 * a^2 * b^6

8. For k = 7: (8 choose 7) * a^(8-7) * b^7 = 8 * a * b^7

9. For k = 8: (8 choose 8) * a^(8-8) * b^8 = 1 * b^8

Now, combine all the terms:

(a + b)^8 = a^8 + 8a^7b + 28a^6b^2 + 56a^5b^3 + 70a^4b^4 + 56a^3b^5 + 28a^2b^6 + 8ab^7 + b^8

So, the expanded form of (a + b)^8 is: a^8 + 8a^7b + 28a^6b^2 + 56a^5b^3 + 70a^4b^4 + 56a^3b^5 + 28a^2b^6 + 8ab^7 + b^8