smc-1088-20-Independent Term

Combinatorics, EXT1 A1 2017 HSC 9 MC

When expanded, which expression has a non-zero constant term?

A. `(x + 1/(x^2))^7`

B. `(x^2 + 1/(x^3))^7`

C. `(x^3 + 1/(x^4))^7`

D. `(x^4 + 1/(x^5))^7`

Show Answers Only

`C`

Show Worked Solution

`text(Consider the general term for option)\ A:`

`T_k`	`= \ ^7C_k · x^(7 – k) · x^(−2k)`
	`= \ ^7C_k · x^(7 – 3k)`

`text(Non zero constant term occurs when)`

`7 – 3k`	`= 0`
`k`	`= 7/3 => text(no terms exists)\ (k\ text{not integer)}`

`text(Consider option)\ C:`

`T_k`	`= \ ^7C_k · x^(3(7 – k)) · x^(−4k)`
	`= \ ^7C_k · x^(21 – 7k)`

`21 – 7k`	`= 0`
`k`	`= 3`

`:.\ text(Non-zero constant term exists)`

`text(since)\ k\ text(is an integer)`

`⇒C`

Combinatorics, EXT1′ A1 2016 HSC 6 MC

Let `p(x) = 1 + x + x^2 + x^3 + … + x^12.`

What is the coefficient of `x^8` in the expansion of `p (x + 1)?`

`1`
`495`
`715`
`1287`

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`=> C`

Show Worked Solution

`p(x + 1) = 1 + (x + 1) + (x + 1)^2 + … + (x + 1)^12`

`text(Coefficient of)\ x^8`

`=\ ^12C_8 +\ ^11C_8 +\ ^10C_8 +\ ^9C_8 +\ ^8C_8`

`= 715`

`=> C`

Combinatorics, EXT1 A1 2005 HSC 2b

Use the binomial theorem to find the term independent of `x` in the expansion of

`(2x - 1/x^2)^12.` (3 marks)

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`((12), (4)) * 2^8 * (-1)^4 = 126\ 720`

Show Worked Solution

`T_k =\ text(General term of)\ \ (2x – 1/x^2)^12`

`T_k`	`= ((12), (k)) (2x)^(12 – k) * (-1)^k * (x^-2)^k`
	`= ((12), (k)) * 2^(12 – k) * x^(12 – k) * (-1)^k * x^(-2k)`
	`= ((12), (k)) * 2^(12 – k) * (-1)^k * x^(12 – 3k)`

`text(Independent term occurs when)`

MARKER’S COMMENT: The general term formula was well known, but many could not apply it to this question.

`x^(12-3k)`	`= x^0`
`12 – 3k`	`= 0`
`k`	`= 4`

`:.\ text(Independent term is)`

`((12), (4)) * 2^8 * (-1)^4 = 126\ 720`

Combinatorics, EXT1 A1 2015 HSC 13b

Consider the binomial expansion

`(2x + 1/(3x))^18 = a_0x^(18) + a_1x^(16) + a_2x^(14) + …`

where `a_0, a_1, a_2`, . . . are constants.

Find an expression for `a_2`. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
Find an expression for the term independent of `x`. (2 marks)

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`(\ ^18C_2 · 2^(16))/(3^2)`
`(\ ^(18)C_9 · 2^9)/(3^9)`

Show Worked Solution

i. `text(Need co-efficient of)\ x^(14)`

`text(General term of)\ (2x + 1/(3x))^(18)`

`T_k`	`= \ ^(18)C_k(2x)^(18 − k) · (1/(3x))^k`
	`= \ ^(18)C_k · 2^(18 − k) · x^(18 − k) · 3^(−k) · x^(−k)`
	`= \ ^(18)C_k · 2^(18 − k) · 3^(−k) · x^(18 − 2k)`

`a_2\ text(occurs when:)`

`18 − 2k`	`= 14`
`2k`	`= 4`
`k`	`= 2`

`:.a_2`	`= \ ^(18)C_2 · 2^(18 − 2) · 3^(−2)`
	`= (\ ^(18)C_2 · 2^(16))/(3^2)`

ii. `text(Independent term occurs when:)`

`18 − 2k`	`= 0`
`2k`	`= 18`
`k`	`= 9`

`:.\ text(Independent term)`

`= \ ^(18)C_9 · 2^(18− 9) · 3^(−9)`

`= (\ ^(18)C_9 · 2^9)/(3^9)`

Combinatorics, EXT1 A1 2014 HSC 3 MC

What is the constant term in the binomial expansion of `(2x - 5/(x^3))^12`?

`((12),(3)) 2^9 5^3`
`((12),(9)) 2^3 5^9`
`-((12),(3)) 2^9 5^3`
`-((12),(9)) 2^3 5^9`

Show Answers Only

`C`

Show Worked Solution

`text(General term)`

`T_k`	`= ((12),(k)) (2x)^(12-k) * (-1)^k *(5x^(-3))^k`
	`= ((12),(k)) 2^(12-k) * x^(12-k) * (-1)^k * 5^k * x^(-3k)`
	`= ((12),(k)) (-1)^k * 2^(12-k) * 5^k * x^(12-4k)`

`text(Constant term when)`

`12 – 4k`	`= 0`
`k`	`= 3`

`:.\ text(Constant term)`

`=((12),(3)) (-1)^3 * 2^9 * 5^3`

`= – ((12),(3)) * 2^9 * 5^3`

`=> C`

Combinatorics, EXT1 A1 2012 HSC 11f

Use the binomial theorem to find an expression for the constant term in the expansion of

`(2x^3 - 1/x)^12`. (2 marks)

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For what values of `n` does `(2x^3 - 1/x)^n` have a non-zero constant term? (1 mark)

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Show Answers Only

`-1760`
`n\ text(must be a multiple of 4)`

Show Worked Solution

i. `text(General term)`

`\ ^12C_k * (2x^3)^(12\ – k) (-1/x)^k`

`=\ ^12C_k * (–1)^k *2^(12\ – k) * x^(36\ – 3k) * x^(-k)`

`=\ ^12C_k * (–1)^k*2^(12\ – k) * x^(36\ – 4k)`

`text(Constant term occurs when)`

`36\ – 4k`	`= 0`
`k`	`= 9`

`:.\ text(Constant term)`	`=\ ^12C_9 * (–1)^9*2^3`
	`= – (12!)/(3!9!) xx 8`
	`= – 1760`

ii. `text(General term of)\ (2x^3\ – 1/x)^n`

♦♦♦ Mean mark 16%.

`\ ^nC_k * (2x^3)^(n\ – k) (–1/x)^k`

`=\ ^nC_k * 2^(n\ – k) * x^(3n\ – 3k) * (–1)^k * x^(-k)`

`=\ ^nC_k * (–1)^k*2^(n\ -k) * x^(3n\ – 4k)`

`text(Constant term when)\ \ 3n\ – 4k = 0.`

`text(i.e.)\ \ k=3/4n`

`text(S)text(ince)\ n\ text(and)\ k\ text(must be integers,)\ \ n\ \ text(must)`

`text(be a multiple of 4.)`