Question

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.

Answer 1

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

Answer 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