Find the coefficient of x^7 for (x-3)^11

Use the binomial theorem to expand (2y-3x)^5

Prove that (n over r)= (n over n-r) for all integers where n is greater than or equal to r and r is greater than or equal to zero

Prove that (n over n-2) + ( n+1 over n-1)= n^2 for all integers n is greater than or equal to 2.

x^7 is the 5th term in the expansion, so it is

11C4 x^7 (-3)^4 = 26730x^7

The coefficients are 1 5 10 10 5 1, so
(2y-3x)^5 =
(2y)^5 + 5(2y)^4(-3x) + 10(2y)^3(-3x)^2 + 10(2y)^2(-3x)^3 + 5(2y)(-3x)^4 + (-3y)^5
Now just expand all those values

nCr = n!/[(n-r)!r!]
nC(n-r) = n!/[(n-(n-r))!(n-r)!)
which you can see is exactly the same.

nCn-2 = nC2 = n(n-1)/2
(n+1)C(n-1) = (n+1)C2 = (n+1)(n)/2
Now subtract and you get n^2

general term(r+1)

= C(11,r) x^(11-r)(-3)^r

so 11-r = 7
r = 4

term(5) = C(11,4) x^7 (-3)^4
= 26730 x^7

making the coefficient 26730

check with Wolfram
http://www.wolframalpha.com/input/?i=%28x-3%29%5E11

(2y-3x)^5
= (2y)^5 + C(5,1)(2y)^4 (-3x) + C(5,2)(2y)^3 (-3x)^2 + C(5,3)(2y)^2 (-3x)^3 + C(5,4)(2y) (-3x)^4 + (-3x)^5
= 32y^5 - 240xy^4 + 720x^2y^3 - 1080x^3y^2 + 810x^4 y - 243x^5

The last two should have factorials in them

The way they sit, they are not true.
eg in the 2nd one:
let n = 3
LS = 3/1 + 4/2 = 3 + 2 = 5
RS = 9
statement is false

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.

To find the coefficient of x^7 for (x-3)^11, we can use the binomial theorem. The binomial theorem states that for any real numbers a and b, and any positive integer n, the expansion of (a + b)^n can be written as the sum of binomial coefficients multiplied by the corresponding powers of a and b.

Using this theorem, we can expand (x-3)^11 as follows:

(x-3)^11 = (11 choose 0) * x^11 * (-3)^0 + (11 choose 1) * x^10 * (-3)^1 + (11 choose 2) * x^9 * (-3)^2 + ... + (11 choose 7) * x^4 * (-3)^7 + ... + (11 choose 11) * x^0 * (-3)^11

The binomial coefficient (n choose r) represents the number of ways to choose r items from a set of n items and is calculated as:

(n choose r) = n! / (r! * (n-r)!)

where ! denotes factorial.

We are interested in the coefficient of x^7, which is obtained by multiplying (11 choose 7) with the appropriate powers of x and (-3):

Coefficient of x^7 = (11 choose 7) * x^7 * (-3)^4

To calculate (11 choose 7), we substitute the values into the formula:

(11 choose 7) = 11! / (7! * (11-7)!)

Simplifying the expression gives:

(11 choose 7) = 11! / (7! * 4!)

(11 choose 7) = (11 * 10 * 9 * 8 * 7!) / (7! * 4 * 3 * 2 * 1)

(11 choose 7) = (11 * 10 * 9 * 8) / (4 * 3 * 2 * 1)

(11 choose 7) = 330

Substituting this value back into the equation for the coefficient of x^7, we have:

Coefficient of x^7 = 330 * x^7 * (-3)^4

Therefore, the coefficient of x^7 for (x-3)^11 is 330 * (-81) = -26730.

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To expand (2y-3x)^5, we can also use the binomial theorem. The logic and steps are similar to the previous example.

The expanded form of (2y-3x)^5 can be calculated as follows:

(2y-3x)^5 = (5 choose 0) * (2y)^5 * (-3x)^0 + (5 choose 1) * (2y)^4 * (-3x)^1 + (5 choose 2) * (2y)^3 * (-3x)^2 + ... + (5 choose 5) * (2y)^0 * (-3x)^5

Simplify the expressions:

(2y-3x)^5 = (2y)^5 + 5 * (2y)^4 * (-3x) + 10 * (2y)^3 * (-3x)^2 + ... + 10 * (2y) * (-3x)^4 + (-3x)^5

Expand and simplify the exponents:

(2y-3x)^5 = 32y^5 - 240y^4x + 720y^3x^2 - 1080y^2x^3 + 810yx^4 - 243x^5

Therefore, the expansion of (2y-3x)^5 is 32y^5 - 240y^4x + 720y^3x^2 - 1080y^2x^3 + 810yx^4 - 243x^5.

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To 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, we can use the formula for binomial coefficients.

The binomial coefficient (n choose r) represents the number of ways to choose r items from a set of n items. It is calculated as:

(n choose r) = n! / (r! * (n-r)!)

Let's consider (n choose r) and (n choose n-r):

(n choose r) = n! / (r! * (n-r)!)
(n choose n-r) = n! / ((n-r)! * (n-(n-r))!)

Simplify the expression for (n choose n-r):

(n choose n-r) = n! / ((n-r)! * (n-n+r)!)
(n choose n-r) = n! / ((n-r)! * r!)

Notice that the only difference between the two expressions is the term in the denominator: (n-r)! vs. r!. However, we can observe that (n-r)! is the same as r! since n-r equals n-n+r.

Therefore, (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.

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To prove that (n choose n-2) + (n+1 choose n-1) = n^2 for all integers n greater than or equal to 2, we can once again use the formula for binomial coefficients.

The equation states:

(n choose n-2) + (n+1 choose n-1) = n^2

We know that (n choose r) = n! / (r! * (n-r)!). Let's substitute the values into the equation:

(n choose n-2) + (n+1 choose n-1) = n^2

(n! / ((n-2)! * (n - (n-2))!)) + ((n+1)! / ((n-1)! * (n+1 - (n-1))!)) = n^2

(n! / ((n-2)! * 2!)) + ((n+1)! / ((n-1)! * 2!)) = n^2

(n! / ((n-2)! * 2)) + ((n+1)! / ((n-1)! * 2)) = n^2

Simplifying the expression:

(n! / ((n-2)! * 2) + ((n+1)! / ((n-1)! * 2)) = n^2

(n * (n-1) / 2) + ((n+1) * n / 2) = n^2

(n^2 - n + n^2 + n) / 2 = n^2

(2n^2) / 2 = n^2

n^2 = n^2

Therefore, we have proved that (n choose n-2) + (n+1 choose n-1) = n^2 for all integers n greater than or equal to 2.