Hard C4 Binomial Expansion Question HELP!

I've been struggling with these questions on the binomial expansion.

a) Find the binomial expansion of (1 + kx)^-2 up to and including the term x³^,where k is a constant.
b) The binomial expansion of (1 + kx)^-2 - (1 + x)ⁿ is 6x²+ px³
(b)(i). Show that n = -2k
(b)(ii) Given that n < 0, find the value of p

I've done part (a) and (b)(i), but no idea how to even start on (b)(ii)! Can someone PLEASE help!!!

Thankyou so much!

Reply 1

TenOfThem

18

Original post by CleverGirl383

I've been struggling with these questions on the binomial expansion.

a) Find the binomial expansion of (1 + kx)^-2 up to and including the term x³^,where k is a constant.
b) The binomial expansion of (1 + kx)^-2 - (1 + x)ⁿ is 6x²+ px³
(b)(i). Show that n = -2k
(b)(ii) Given that n < 0, find the value of p

I've done part (a) and (b)(i), but no idea how to even start on (b)(ii)! Can someone PLEASE help!!!

Thankyou so much!

Use the 6x^2 to find k and n

Reply 2

JSShaw

you need to find the coefficient of x^3 you do this by finding the cofficient of x^3 in both (1+kx)^-2 and -(1+x)^n. CARE FOR THE MINUS. the "given that n<0" means the exponent of the binomial expansion is negative and therefore goes on forever, telling the person solving it that it has an expansion for 3!n(n-1)(n-2)x^3

Reply 3

CleverGirl383

OP

5

Original post by JSShaw

you need to find the coefficient of x^3 you do this by finding the cofficient of x^3 in both (1+kx)^-2 and -(1+x)^n. CARE FOR THE MINUS. the "given that n<0" means the exponent of the binomial expansion is negative and therefore goes on forever, telling the person solving it that it has an expansion for 3!n(n-1)(n-2)x^3

I've got the coefficient of x³ in (1 + kx)^-2 as -4k³x³ and as for (1 + x)ⁿ I got [(n³ - 3n² + 2n)x³] / 6 but don't know what I'm supposed to do with this?

Reply 4

JSShaw

Original post by CleverGirl383

I've got the coefficient of x³ in (1 + kx)^-2 as -4k³x³ and as for (1 + x)ⁿ I got [(n³ - 3n² + 2n)x³] / 6 but don't know what I'm supposed to do with this?

Read the question, you have an equation from the previous question for possibly substituting

Reply 5

WhiteGroupMaths

9

Original post by CleverGirl383

I've been struggling with these questions on the binomial expansion.

a) Find the binomial expansion of (1 + kx)^-2 up to and including the term x³^,where k is a constant.
b) The binomial expansion of (1 + kx)^-2 - (1 + x)ⁿ is 6x²+ px³
(b)(i). Show that n = -2k
(b)(ii) Given that n < 0, find the value of p

I've done part (a) and (b)(i), but no idea how to even start on (b)(ii)! Can someone PLEASE help!!!

Thankyou so much!

For starters, have you obtained the coefficient of x² in the expansion of (1 + kx)^-2 - (1 + x)ⁿ? Set that to be equal to 6 by comparison, and see if you can arrive at n=-2k.

Peace.

(edited 9 years ago)

Reply 6

TenOfThem

18

Original post by CleverGirl383

I've got the coefficient of x³ in (1 + kx)^-2 as -4k³x³ and as for (1 + x)ⁿ I got [(n³ - 3n² + 2n)x³] / 6 but don't know what I'm supposed to do with this?

I repeat

Use 6x^2 to find n and k

Reply 7

TenOfThem

18

Original post by WhiteGroupMaths

For starters, have you obtained the coefficient of x² in the expansion of (1 + kx)^-2 - (1 + x)ⁿ? Set that to be equal to 6 by comparison, and see if you can arrive at n=-2k.

Peace.

She will have used the coefficient of x to get that result