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# another binomial q watch

1. edexcel p2 jan 02 q 9

f(x)=(1+x/k)n n>2 k,n E N

the coefficient of x3 is twice the coefficient of x2 in the binomail explansion of f(x)

prove that n=6k+2

so I said

2 n!/2(n-2)!k2 =n!/6(n-3)(n-2)!k3

cancelling out gives

1/k2 =1/6(n-3)k3

then 1=(6n-18)k

then I get stuck and cant rearrange to get n=

Also the mark scheem is very different

they give

2n(n-1)/2k2 =n(n-1)(n-2)/6k3

6k=n-2
n=6k+2

2. 2 n!/2(n-2)!k2 =n!/6(n-3)(n-2)!k3
This is wrong. ^^

The correct expansion is:

(1 + x/k)n = 1 + n(x/k) + [n(n-1)(x/k)2]/2 + [n(n-1)(n-2)(x/k)3/6 ...

So: 2[n(n-1)(1/k)2]/2 = [n(n-1)(n-2)(1/k)3/6

=> 2/2 = (n - 2)/6k

=> 6k = n - 2

=> n = 6k + 2
3. Yeah i kinda dont understand why ur using factorials everywhere??

f(x)=(1+x/k)n

f(x)= 1 + nx/k + (n(n-1)(x/k)²)/(2!) + (n(n-1)(n-2)(x/k)³)/(3!)+.....+

if 2x² = x³ coeff then
2(n(n-1)(1/k)²)/(2!) = (n(n-1)(n-2)(1/k)³)/(3!)
1/2k²= (n-2)/6k³
6k = n-2
n=6k+2
4. (Original post by samdavyson)
This is wrong. ^^

The correct expansion is:

(1 + x/k)n = 1 + n(x/k) + [n(n-1)(x/k)2]/2 + [n(n-1)(n-2)(x/k)3/6 ...

So: 2[n(n-1)(1/k)2]/2 = [n(n-1)(n-2)(1/k)3/6

=> 2/2 = (n - 2)/6k

=> 6k = n - 2

=> n = 6k + 2
where does the correct expansion come from, i don't understand where the (n-1) come from.

I was taught to solve using factorials to change n c r into n!/r!(n-r)!

so they come from the nc2 and nc3

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