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C2 Question

The coefficient of x²
in the binomial expansion of (1 + 0.4x)ⁿ
, where n is a positive integer, is 1.6

a Find the value of n.
b Use your value of n to find the coefficient of x
4
in the expansion.

need help on a, have no idea where to start,
thanks =)

Reply 1

Noble.

Firstly, do you know how to work out the binomial expansion?

Reply 2

Mr M

Study Forum Helper

Expand

(1+0.4x)^n

as far as the term in

x^2

Set the coefficient of

x^2

equal to 1.6.

Reply 3

NidaJaffri

Compare coefficient of the x^2 term

Reply 4

Hi, How are you ?

Original post by Noble.

Firstly, do you know how to work out the binomial expansion?

yes, using NcR notation.

Reply 5

Noble.

Original post by Hi, How are you ?

yes, using NcR notation.

Ok, Mr. M has said how to proceed once you've done the binomial expansion up to the

x^2

term.

Reply 6

Hi, How are you ?

Original post by Mr M

Expand

(1+0.4x)^n

as far as the term in

x^2

Set the coefficient of

x^2

equal to 1.6.

Tried using NcR notation, and then I got

\displaystyle \binom{n}{0}

^ 1ⁿ +

\displaystyle \binom{n}{1}

^ 1^n-1 ^ 0.4x +

\displaystyle \binom{n}{2}

^ 1^n-2 ^ (0.4x)²
, What do i do next ?

Reply 7

Noble.

(1+x)^n = {n \choose 0}x^0 + {n \choose 1}x^1 + {n \choose 2}x^2 + \cdots + {n \choose {n-1}}x^{n-1} + {n \choose n}x^n

(1+0.4x)^n = {n \choose 0}(0.4x)^0 + {n \choose 1}(0.4x)^1 + {n \choose 2}(0.4x)^2 + \cdots

You know that

{n \choose 2}(0.4x)^2 = 1.6x^2

You can use this to determine

n

Reply 8

Hi, How are you ?

Original post by Noble.

(1+x)^n = {n \choose 0}x^0 + {n \choose 1}x^1 + {n \choose 2}x^2 + \cdots + {n \choose {n-1}}x^{n-1} + {n \choose n}x^n

(1+0.4x)^n = {n \choose 0}(0.4x)^0 + {n \choose 1}(0.4x)^1 + {n \choose 2}(0.4x)^2 + \cdots

You know that

{n \choose 2}(0.4x)^2 = 1.6x^2

You can use this to determine

n

How would you get the n out of the brackets, what process would you do??

Reply 9

Mr M

Study Forum Helper

Original post by Hi, How are you ?

How would you get the n out of the brackets, what process would you do??

\displaystyle \binom{n}{r} = \frac{n!}{r!(n-r)!}

but the easiest way is just to write down a few rows of Pascal's triangle!

Reply 10

Noble.

Original post by Hi, How are you ?

How would you get the n out of the brackets, what process would you do??

By definition

{n \choose m} = \dfrac{n!}{m!(n-m)!}

Reply 11

Hi, How are you ?

Original post by Noble.

By definition

{n \choose m} = \dfrac{n!}{m!(n-m)!}

so it means that

\displaystyle \binom{n}{2}

= n! / 2!(n-2)!
= n! / 2(n-2)!

Reply 12

Noble.

Original post by Hi, How are you ?

so it means that

\displaystyle \binom{n}{2}

= n! / 2!(n-2)!
= n! / 2(n-2)!

Yep. Now note that n! = (n-1)(n-2)(n-3) x ... x 1 and (n-2)! = (n-2)(n-3)(n-4) x... x 1

So you should be able to see what terms cancel in the fraction.

Reply 13

Mr M

Study Forum Helper

Original post by Noble.

Yep. Now note that n! = (n-1)(n-2)(n-3) x ... x 1 and (n-2)! = (n-2)(n-3)(n-4) x... x 1

So you should be able to see what terms cancel in the fraction.