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Prove by mathematical induction that $8^{2n+1} + 6^{2n-1}$ is divisible by 7, for any integer $n e 1$ - HSC - SSCE Mathematics Extension 1 - Question 14 - 2017 - Paper 1
Question 14
Prove by mathematical induction that $8^{2n+1} + 6^{2n-1}$ is divisible by 7, for any integer $n e 1$. Let $P(n)$ be the given proposition. $P(1)$ is true since $... show full transcript
Worked Solution & Example Answer:Prove by mathematical induction that $8^{2n+1} + 6^{2n-1}$ is divisible by 7, for any integer $n e 1$ - HSC - SSCE Mathematics Extension 1 - Question 14 - 2017 - Paper 1
Step 1
Prove by mathematical induction that $8^{2n+1} + 6^{2n-1}$ is divisible by 7, for any integer $n e 1$
Answer
Let ( P(n) ) be the given proposition.
( P(1) ) is true since ( 8^2 \times 6^0 = 7 \times 4 ) which is divisible by 7.
Let ( k ) be an integer for which ( P(k) ) is true. That is ( 8^{2k+1} + 6^{2k-1} = 7m ), for some integer ( m ).
Consider ( 8^{2(k+1)+1} + 6^{2(k+1)-1} ):
( 8^{2(k+1)+1} + 6^{2(k+1)-1} = 8^{2k+3} + 6^{2k+1} )
This can be rewritten as:
( = 8^{2k+1} \times 8^2 + 6^{2k-1} \times 6^2 )
Substituting for ( 8^{2k+1} ) and ( 6^{2k-1} ) using ( P(k) ), we get:
( = 7m \times 64 + 7n \times 36 )
Which simplifies to:
( = 7 \times (64m + 36n) )
Thus, ( P(k+1) ) is true.
By induction, ( P(n) ) holds for all ( n \geq 1 ).
Step 2
Show that the $x$ coordinates of $R$ and $Q$ satisfy: $x^2 + 4apx - 4p^2 = 0$
Answer
The sum of the roots of the equation is (-4ap).
Let the -coordinates of and be ( x_1 ) and ( x_2 ). Thus:
( x_1 + x_2 = -4ap ) ( x_1x_2 = -4p^2 )
This leads to the quadratic equation ( x^2 + 4apx - 4p^2 = 0 ).
Step 3
Show that the coordinates of $M$ are $(-2ap,-p^2(2a+1))$
Answer
Substituting into the tangent equation at point :
( y = px - p^2 )
The average of the roots gives us the -coordinate of :
( M = -2ap )
For the -coordinate:
( y = -p^2(2a+1) )
Thus, the coordinates of become ( (-2ap, -p^2(2a+1)) ).
Step 4
Find the value of $a$ so that the point $M$ always lies on the parabola $x^2 = -4y$
Answer
To ensure point lies on the parabola, substitute into ( x^2 = -4y ):
( (-2ap)^2 = -4(-p^2(2a + 1)) )
This simplifies to:
( 4a^2p^2 = 4p^2(2a + 1) )
Canceling ( 4p^2 ) (assuming ), we have:
( a^2 = 2a + 1 )
Solving gives: ( (a - 1)^2 = 0 ) hence ( a = 1 ).
Step 5
By differentiating the product $F(t)e^{0.4t}$ show that $ rac{d}{dt}(F(t)e^{0.4t}) = 50e^{-0.1t}$
Answer
Using the product rule, we differentiate:
( rac{d}{dt}(F(t)e^{0.4t}) = F'(t)e^{0.4t} + F(t)(0.4e^{0.4t}) )
Substituting for ( F'(t) ), we get:
( F'(t)e^{0.4t} = (50e^{-0.5t} - 0.4F(t))e^{0.4t} )
This simplifies to:
( = 50e^{-0.1t} ).
Step 6
Hence, or otherwise, show that $F(t) = 500(e^{-0.4t}-e^{-0.5t})$
Answer
Integrating the rate of change:\n ( F(t)e^{0.4t} = \int 50e^{-0.1t}dt )
This gives us:
( F(t)e^{0.4t} = -500e^{-0.1t} + C )
Setting the initial condition ( F(0) = 0 ), we find ( C = 500 ). Thus:
( F(t) = 500(e^{-0.4t} - e^{-0.5t}) ).
Step 7
For what value of $t$ does this maximum occur?
Answer
To find the maximum, we need to set the derivative ( F'(t) = 0 ):
( 50e^{-0.4t} - 200e^{-0.5t} = 0 )
This leads to:
( e^{-0.4t} = 4e^{-0.5t} )
Taking logarithms, we deduce:
( t = 10 \ln(4) )
Approximating gives ( t \approx 2.23 \text{ hours} ).