smc-5115-60-Recursive

Proof, EXT2 P2 EQ-Bank 11

A sequence is given by the recursive formula

`a_1=10, \ a_(n+1)=3a_n+4` for `n>=1`

Using mathematical induction to show the formula for the general term of the sequence is

`a_n=4(3^n)-2` (3 marks)

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`text{Proof (See Worked Solution)}`

Show Worked Solution

`text{Prove true for}\ \ n=1:`

`text{LHS}\ =a_1=10`

`text{RHS}\ =4(3^1)-2=10=\ text{LHS}`

`:.\ text{True for}\ \ n=1`

`text{Assume true for}\ \ n=k:`

`a_k=4(3^k)-2\ \ text{… (1)}`

`text{Prove true for}\ \ n=k+1:`

`text{i.e.}\ a_(k+1)=4(3^(k+1))-2`

`a_(k+1)`	`=3a_k+4`
	`=3[4(3^k)-2]+4`
	`=43^k3-6+4`
	`=4(3^(k+1))-2`
	`=\ text{RHS}`

`:.\ text{True for}\ \ n=k+1`

`:.\ text{Since true for} \ n=1,\ text{by PMI, true for integers} \ n>=1.`

Calculus, EXT2 C1 2022 HSC 14b

Let `J_(n)=int_(0)^(1)x^(n)e^(-x)\ dx`, where "n" is a non-negative integer.

Show that `J_(0)=1-(1)/(e)`. (1 mark)

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Show that `J_(n) <= (1)/(n+1)`. (2 marks)

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Show that `J_(n)=nJ_(n-1)-(1)/(e)`. (2 marks)

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Using parts (i) and (iii), show by mathematical induction, or otherwise, that for all `n >= 1`,
`J_(n)=n!-(n!)/(e)sum_(r=0)^(n)(1)/(r!)` (2 marks)

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Using parts (ii) and (iv) prove that `e=lim_(n rarr oo)sum_(r=0)^(n)(1)/(r!)`. (1 mark)

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`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`

Show Worked Solution

i.	`J_0`	`=int_0^1 e^(-x)\ dx`
		`=[-e^(-x)]_0^1`
		`=-e^(-1)+1`
		`=1-1/e`

Mean mark (i) 93%.

ii. `text{Show}\ \ J_n<=1/(n+1)`

`text{Note:}\ e^(-x)<1\ \ text{for}\ \ x in [0,1]`

`J_n`	`=int_0^1 x^n e^(-x)\ dx`
	`leq int_0^1 x^n \ dx`
	`leq 1/(n+1)[x^(n+1)]_0^1`
	`leq 1/(n+1)(1^(n+1)-0)`
	`leq 1/(n+1)\ \ text{… as required}`

♦♦ Mean mark (ii) 28%.

iii. `text{Show}\ \ J_n=nJ_(n-1)-1/e`

`u`	`=x^n`	`v′`	`=e^(-x)`
`u′`	`=nx^(n-1)`	`v`	`=-e^(-x)`

`J_n`	`=[-x^n * e^(-x)]_0^1-int_0^1 nx^(n-1)*-e^(-x)\ dx`
	`=(-1^n * e^(-1)+0^n e^0)+nint_0^1 x^(n-1)*e^(-x)\ dx`
	`=nJ_(n-1)-1/e`

iv. `text{Prove}\ \ J_(n)=n!-(n!)/(e)sum_(r=0)^(n)(1)/(r!)\ \ text{for}\ \ n >= 0`

`text{If}\ \ n=0:`

`text{LHS} = 1-1/e\ \ text{(see part (i))}`

`text{RHS} = 0!-0!/e (1/(0!)) = 1-1/e(1)=\ text{LHS}`

`:.\ text{True for}\ \ n=0.`

`text{Assume true for}\ \ n=k:`

`J_(k)=k!-(k!)/(e)sum_(r=0)^(k)(1)/(r!)`

♦ Mean mark (iv) 50%.

`text{Prove true for}\ \ n=k+1:`

`text{i.e.}\ \ J_(k+1)=(k+1)!-((k+1!))/(e)sum_(r=0)^(k+1)(1)/(r!)`

`J_(k+1)`	`=(k+1)J_k-1/e\ \ text{(using part (iii))}`
	`=(k+1)(k!-(k!)/(e)sum_(r=0)^(k)(1)/(r!))-1/e`
	`=(k+1)!-((k+1)!)/(e)sum_(r=0)^(k)(1)/(r!)-1/e xx ((k+1)!)/((k+1)!)`
	`=(k+1)!-((k+1)!)/e(\ sum_(\ r=0)^(k)(1)/(r!)+1/((k+1)!))`
	`=(k+1)!-((k+1)!)/e(\ sum_(\ r=0)^(k+1)(1)/(r!))`

`=>\ text{True for}\ \ n=k+1`

`:.\ text{S}text{ince true for}\ n=1,\ text{by PMI, true for integers}\ n>=1`

v. `0<=J_n<= 1/(n+1)\ \ \ text{(part (ii))}`

`lim_(n->oo) 1/(n+1)=0\ \ => \ lim_(n->oo) J_n=0`

`text{Using part (iv):}`

`J_n/(n!)`	`=1-1/e sum_(r=0)^(n)(1)/(r!)`
`1/e sum_(r=0)^(n)(1)/(r!)`	`=1-J_n/(n!)`
`sum_(r=0)^(n)(1)/(r!)`	`=e-(eJ_n)/(n!)`
`lim_(n->oo)(\ sum_(\ r=0)^(n)(1)/(r!))`	`=lim_(n->oo)(e-(eJ_n)/(n!))`
	`=e-0`
	`=e`

♦♦ Mean mark (v) 34%.

Proof, EXT2 P2 2022 HSC 13b

The numbers `a_(n)`, for integers `n >= 1`, are defined as

`{:[a_(1)=sqrt2],[a_(2)=sqrt(2+sqrt2)],[a_(3)=sqrt(2+sqrt(2+sqrt2)) \ text{, and so on.}]:}`

These numbers satisfy the relation `a_(n+1)^(2)=2+a_(n)`, for `n >= 1`. (Do NOT prove this)

Use mathematical induction to prove that `a_(n)=2cos\ pi/(2^(n+1))`, for all integers `n >= 1`. (4 marks)

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`text{Proof (See Worked Solution)}`

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`text{Prove}\ \ a_(n)=2cos(pi/(2^(n+1)))\ \ text{for}\ \ n >= 1`

`text{If}\ \ n=1:`

`a_1=2cos((pi)/(2^2))=2xx1/sqrt2=sqrt2`

`:.\ text{True for}\ n=1.`

Mean mark 58%.

`text{Assume true for}\ \ n=k:`

`a_(k)= =2cos(pi/(2^(k+1)))`

`text{Prove true for}\ \ n=k+1:`

`text{i.e.}\ \ a_(k+1)= 2cos(pi/(2^(k+2)))`

`text{LHS}`	`=sqrt(2+a_k)`
	`=sqrt(2+2cos(pi/(2^(k+1)))`
	`=sqrt(2+2 cos(2 xx (pi/(2^(k+2)))))\ \ \ text{(using}\ \ cos(2theta)=2cos^2theta-1text{)}`
	`=sqrt(2+2(2cos^2(pi/(2^(k+2)))-1)`
	`=sqrt(2+4cos^2(pi/(2^(k+2)))-2)`
	`=sqrt(4cos^2(pi/(2^(k+2)))`
	`=2cos(pi/(2^(k+2)))`
	`=\ text{RHS}`

Proof, EXT2 P2 2006 HSC 7c

The sequence `{x_n}` is given by

`x_1 = 1` and `x_(n + 1) = (4 + x_n)/(1 + x_n)` for `n >= 1.`

Prove by induction that for `n >= 1`, `x_n = 2 ((1 + alpha^n)/(1 - alpha^n))`, where `alpha = -1/3.` (4 marks)

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Hence find the limiting value of `x_n` as `n -> oo.` (1 mark)

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`text(Proof)\ \ text{(See Worked Solutions)}`
`2`

Show Worked Solution

i. `text(If)\ \ n = 1`

`x_1=2 ((1 – 1/3)/(1 + 1/3))=(2 xx 2/3)/(4/3)=1`

`:.text(True for)\ \ n = 1`

`text(Assume true for)\ \ n=k,`

`text(i.e.)\ \ \ \ x_k = 2 ((1 + alpha^k)/(1 – alpha^k))`

`text(Prove true for)\ \ n=k+1`

`text(i.e.)\ \ \ \ x_(k + 1) = 2 ((1 + alpha^(k + 1))/(1 – alpha^(k+1)))`

`x_(k + 1) =`	`(4 + x_k)/(1 + x_k)`
`=`	`(4 + 2 ((1 + alpha^k)/(1 – alpha^k)))/(1 + 2((1 + alpha^k)/(1 – alpha^k)))`
`=`	`2 [(2(1 – alpha^k) + (1 + alpha^k))/(1 – alpha^k + 2 (1 + alpha^k))]`
`=`	`2 [(2 – 2 alpha^k + 1 + alpha^k)/(1 – alpha^k + 2 + 2 alpha^k)]`
`=`	`2 [(3 – alpha^k)/(3 + alpha^k)]`
`=`	`2 [(1 – alpha^k xx 1/3)/(1 + alpha^k xx 1/3)]`
`=`	`2 [(1 + alpha^k xx alpha)/(1 – alpha^k xx alpha)],\ \ \ \ (alpha=-1/3)`
`=`	`2 [(1 + alpha^(k + 1))/(1 – alpha^(k + 1))]`
`=`	`text(RHS)`

`:.text(True for)\ \ n = k + 1\ \ text(if it is true for)\ \ n = k.`

`:.text(S)text(ince true for)\ \ n=1, text(by PMI, true for all integral)\ \ n >= 1.`

ii. `text(S)text(ince)\ \ lim_(n -> oo) (-1/3)^n=0`

`:.lim_(n -> oo) x_n`	`=2 ((1 + (-1/3)^n)/(1 – (-1/3)^n))`
	`=2`

Proof, EXT2 P2 2010 HSC 6b

A sequence `a_n` is defined by

`a_n = 2a_(n − 1) + a_(n − 2)`,

for `n ≥ 2`, with `a_0 = a_1 = 2`.

Use mathematical induction to prove that

`a_n = (1 + sqrt2)^n + (1 − sqrt2)^n` for all `n ≥ 0`. (3 marks)

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`text{Proof (See Worked Solutions)}`

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`text(Prove)\ \ a_n = (1 + sqrt2)^n + (1 − sqrt2)^n\ text(for all)\ n ≥ 0.`

`text(where)\ a_0 = a_1 = 2, and \ a_n = 2a_(n − 1) + a_(n − 2)\ \ text(for)\ \ n ≥ 2,\ `

`text(When)\ \ n=0,`	`a_0`	`= (1 + sqrt2)^0 + (1 − sqrt2)^0`
		`= 1 + 1=2`
`text(When)\ \ n=1,`	`a_1`	`= (1 + sqrt2)^1 + (1 − sqrt2)^1`
		`= 1 + sqrt2 + 1 − sqrt2=2`

`:.\ text(True for)\ n=0\ \ and\ n=1`

♦ Mean mark 36%.

`text(Assume that)`

`a_k = (1 + sqrt2)^k + (1 − sqrt2)^k\ text(and)`

`a_(k − 1) = (1 + sqrt2)^(k − 1) + (1 − sqrt2)^(k − 1)\ text(for all)\ k ≥ 1.`

`text(Prove true for)\ \ n = k + 1`

`text(i.e.)\ \ a_(k + 1) = (1 + sqrt2)^(k + 1) + (1 − sqrt2)^(k + 1).`

`a_(k + 1)`	`= 2a_k + a_(k − 1).`
	`= 2(1 + sqrt2)^k + 2(1 − sqrt2)^k + (1 + sqrt2)^(k − 1) + (1 − sqrt2)^(k − 1)`
	`= (1 + sqrt2)^(k − 1)(2 + 2sqrt2 + 1) + (1 − sqrt2)^(k − 1)(2 − 2sqrt2 + 1)`
	`= (1 + sqrt2)^(k − 1)(1 + sqrt2)^2 + (1 − sqrt2)^(k − 1)(1− sqrt2)^2`
	`= (1 + sqrt2)^(k + 1) + (1 − sqrt2)^(k + 1)`

`=>\ text(True for)\ \ n = k + 1\ \ text(if it is true for)\ \ n = k\ \ text(and)\ \ n = k − 1.`

`:.\ text(S)text(ince true for)\ n = 0 and n=1,\ text(by PMI, it is)`

`text(true for integral)\ n ≥ 0.`