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Proof, EXT2 P1 2018 HSC 15c

Let  `n`  be a positive integer and let  `x`  be a positive real number.

  1.  Show that  `x^n - 1 - n(x - 1) = (x - 1)(1 + x + x^2 + … + x^(n - 1) - n)`.  (1 mark)

  2.  Hence, show that  `x^n >= 1 + n(x - 1)`.  (2 marks)

  3.  Deduce that for positive real numbers `a` and `b`,
     
          `a^nb^(1-n)>=na + (1-n)b`  (2 marks)

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  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `text(See Worked Solutions)`
Show Worked Solution
i.    `text(RHS)` `= (x – 1)underbrace{(1 + x + x^2 + … + x^(n – 1) – n)}_{text(GP where)\ \ a = 1,\ r = x}`
    `= (x – 1) ((1(x^n – 1))/(x – 1) – n)`
    `= x^n – 1 – n(x – 1)`
    `=\ text(LHS)`

 

ii.   `text(Let)\ P(x) = (x – 1)(1 + x + x^2 + … + x^(n – 1) – n)`

♦♦♦ Mean mark 11%.

`text(If)\ \ x = 1, P(x) = 0`
 

`text(If)\ \ 0 < x < 1, \ (x – 1) < 0, \ (1 + x + x^2 + … + x^(n – 1) – n) < 0`

`=> P(x) > 0`
 

`text(If)\ \ x > 1, \ (x – 1) > 0, \ (1 + x + … + x^(n – 1) – n) > 0`

`=> P(x) > 0`
 

`x^n – 1 – n(x – 1) >= 0`

`:. x^n >= 1 + n(x – 1)`

 

iii.   `x^n >= 1 + n(x – 1)\ \ text(for)\ \ x ∈ R^+`

♦♦ Mean mark 23%.

`text(S)text(ince)\ a,b ∈ R^+`

`(a/b)^n` `>= 1 + n(a/b – 1)`
`(a^n)/(b^n) xx b` `>= b + na – nb,\ \ \ \ (b > 0)`
`a^n b^(1 – n)` `>= na + (1 – n)b`

Proof, EXT2 P1 2014 HSC 16b

Suppose `n` is a positive integer. 

  1. Show that 
     
        `-x^(2n) ≤ 1/(1 + x^2) − (1 − x^2 + x^4 − x^6 + … + (-1)^(n − 1) x^(2n − 2)) ≤ x^(2n)`.  (3 marks)

  2. Use integration to deduce that
     
    `-1/(2n + 1) ≤ pi/4 − (1 − 1/3 + 1/5 − … + (-1)^(n − 1) 1/(2n − 1)) ≤ 1/(2n + 1)`.  (2 marks)

  3. Explain why   `pi/4 = 1 − 1/3 + 1/5 − 1/7 + …`.  (1 mark)

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  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
  3. `text{Proof (See Worked Solutions)}`
Show Worked Solution

i.   `text(Let)\ \ S_n=1 − x^2 + x^4 − x^6 + … + (-1)^(n − 1) x^(2n − 2)`

`=>text(GP where)\ \ a=1, r=-x^2`

♦♦ Mean mark 12%.
STRATEGY: Applying the GP formula and simplifying the middle term is worth 2 full marks in this question.
`:.S_n` `=(1(1-(-x^2)^n))/(1-(-x^2))`
  `= (1 − (-x^2)^n)/(1 + x^2)`

 

`:.1/(1 + x^2) − (1 − x^2 + x^4 − x^6 + … + (-1)^(n − 1) x^(2n − 2))​`

`=1/(1 + x^2)-(1 − (-x^2)^n)/(1 + x^2)`

`=((-x^2)^n)/(1 + x^2)`

`=((-1)^nx^(2n))/(1 + x^2)`

 

`text(S)text(ince)\ \ (1 + x^2)>=1,\ \ \ -x^(2n) ≤ ((-1)^nx^(2n))/(1 + x^2) ≤ x^(2n)`

`:.\ text(We can conclude)`

`-x^(2n) ≤ 1/(1 + x^2) − (1 − x^2 + x^4 − x^6 + … + (-1)^(n − 1) x^(2n − 2))\ ≤ x^(2n)` 

 

ii.   `text{Integrating part (i) between 0 and 1}`

♦ Mean mark 43%.
`int_0^1 -x^(2n)\ dx` `=(-1)/(2n + 1)[x^(2n + 1)]_0^1`
  `=(-1)/(2n + 1)`
`int_0^1 1/(1 + x^2)` `=[tan^(-1) x]_0^1`
  `=pi/4`
`int_0^1 (1 − x^2 + x^4 − x^6 + … + (-1)^(n − 1) x^(2n − 2))\ dx` 
`=[x − (x^3)/3 + (x^5)/5 −… + ((-1)^(n − 1)x^(2n − 1))/(2n − 1)]_0^1`
`=1 − 1/3 + 1/5 − … + ((-1)^(n − 1))/(2n − 1)`
`int_0^1 x^(2n)\ dx` `=1/(2n + 1)[x^(2n + 1)]_0^1`
  `=1/(2n + 1)`

 
`:.\ text(We can conclude)`

`(-1)/(2n + 1) ≤ pi/4 − (1 − 1/3 + 1/5 − … + (-1)^(n − 1) 1/(2n − 1)) ≤ 1/(2n + 1)`

 

iii.  `(-1)/(2n + 1) ≤ pi/4 − (1 − 1/3 + 1/5 − … + ((-1)^(n − 1))/(2n − 1)) ≤ 1/(2n + 1)`

`text(As)\ n → ∞,`

`=>(-1)/(2n + 1) → 0^-\ \ text(and)\ \ 1/(2n + 1) → 0^+`

`=> pi/4 − (1 − 1/3 + 1/5 − … + 1/(2n − 1)) → 0`

♦♦ Mean mark 28%.

`:. pi/4 = 1 -1/3 + 1/5 − 1/7 + …`