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Combinatorics, EXT1 A1 2017 HSC 9 MC

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

  1. `(x + 1/(x^2))^7`
  2. `(x^2 + 1/(x^3))^7`
  3. `(x^3 + 1/(x^4))^7`
  4. `(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(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 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. `1`
  2. `495`
  3. `715`
  4. `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)

Show Answers Only

`((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:}`

`x^(12-3k)= x^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.

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

  2. Find an expression for the term independent of `x`.  (2 marks)

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a.    `(\ ^18C_2 · 2^(16))/(3^2)`

b.    `(\ ^(18)C_9 · 2^9)/(3^9)`

Show Worked Solution

a.     `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\ \ =>\ \ k=2`

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

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

`18-2k= 0\ \ =>\ \ 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`?

  1. `((12),(3)) 2^9 5^3`
  2. `((12),(9)) 2^3 5^9`
  3. `-((12),(3)) 2^9 5^3`
  4. `-((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 occurs 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

  1. Use the binomial theorem to find an expression for the constant term in the expansion of
  2.     `(2x^3-1/x)^12`.   (2 marks)

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

Show Answers Only

a.    `-1760`

b.    `n\ text(must be a multiple of 4)`

Show Worked Solution

a.    `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`
 

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

♦♦♦ Mean mark (ii) 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 occurs when:)`

`3n-4k = 0\ \ =>\ \ k=3/4 n` 

`text(S)text(ince)\ n\ text(and)\ k\ text(must be integers,)\ n\ text(must be a multiple of 4.)`