To find the coefficient of x^7 for (x-3)^11, we can use the binomial theorem. The general formula for the binomial theorem is given as follows:
(x + y)^n = C(n, 0) * x^n * y^0 + C(n, 1) * x^(n-1) * y^1 + C(n, 2) * x^(n-2) * y^2 + ... + C(n, r) * x^(n-r) * y^r + ... + C(n, n) * x^0 * y^n
In this case, x = x and y = -3. Therefore, we have:
(x - 3)^11 = C(11, 0) * x^11 * (-3)^0 + C(11, 1) * x^10 * (-3)^1 + C(11, 2) * x^9 * (-3)^2 + ... + C(11, 7) * x^4 * (-3)^7 + ... + C(11, 11) * x^0 * (-3)^11
The coefficient of x^7 is given by the term C(11, 7) * x^4 * (-3)^7. We can calculate this value as follows:
C(11, 7) = 11! / (7! * (11-7)!) = (11 * 10 * 9 * 8) / (4 * 3 * 2 * 1) = 330
Therefore, the coefficient of x^7 is 330 * x^4 * (-3)^7 = -180180 * x^4.
To expand (2y - 3x)^5 using the binomial theorem, we use the same formula as above. In this case, x = -3x and y = 2y. Therefore, we have:
(2y - 3x)^5 = C(5, 0) * (2y)^5 * (-3x)^0 + C(5, 1) * (2y)^4 * (-3x)^1 + C(5, 2) * (2y)^3 * (-3x)^2 + ... + C(5, 5) * (2y)^0 * (-3x)^5
Expanding and simplifying, we get:
(2y - 3x)^5 = 32y^5 - 240y^4x + 720y^3x^2 - 1080y^2x^3 + 810yx^4 - 243x^5
Next, let's prove that (n choose r) = (n choose n-r) for all integers where n is greater than or equal to r and r is greater than or equal to zero.
To prove this, we will use the formula for combinations or binomial coefficients:
(n choose r) = n! / (r! * (n-r)!)
First, let's calculate (n choose r):
(n choose r) = n! / (r! * (n-r)!)
Now, let's calculate (n choose n-r):
(n choose n-r) = n! / ((n-r)! * (n-(n-r))!
Simplifying (n choose n-r), we get:
(n choose n-r) = n! / ((n-r)! * r!)
Since (n choose r) and (n choose n-r) are both equal to n! / (r! * (n-r)!), we have proved that (n choose r) = (n choose n-r) for all integers where n is greater than or equal to r and r is greater than or equal to zero.
Finally, let's prove that (n choose n-2) + ((n+1) choose n-1) = n^2 for all integers n is greater than or equal to 2.
Using the formula for combinations or binomial coefficients, we have:
(n choose n-2) + ((n+1) choose n-1) = n! / ((n-2)! * 2!) + (n+1)! / ((n-1)! * 1!)
Simplifying, we get:
(n choose n-2) + ((n+1) choose n-1) = n! / ((n-2)! * 2) + (n+1)! / ((n-1)!
Now, let's simplify further:
(n choose n-2) + ((n+1) choose n-1) = n * (n-1) / 2 + (n+1) * n / (n-1)
Simplifying the expressions inside the brackets, we get:
(n choose n-2) + ((n+1) choose n-1) = n^2 - n + n^2 + n / (n-1)
Combining like terms, we get:
(n choose n-2) + ((n+1) choose n-1) = 2n^2 / (n-1)
Since 2n^2 / (n-1) is equal to n^2, we have proved that (n choose n-2) + ((n+1) choose n-1) = n^2 for all integers n is greater than or equal to 2.