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Use mathematical induction to prove that, for n \geq 1, 1 \times 5 + 2 \times 6 + 3 \times 7 + \cdots + n( n + 13) = \frac{1}{6} n(n + 1)(2n + 13) - HSC - SSCE Mathematics Extension 1 - Question 6 - 2011 - Paper 1
Question 6
Use mathematical induction to prove that, for n \geq 1, 1 \times 5 + 2 \times 6 + 3 \times 7 + \cdots + n( n + 13) = \frac{1}{6} n(n + 1)(2n + 13). The diagram sho... show full transcript
Worked Solution & Example Answer:Use mathematical induction to prove that, for n \geq 1, 1 \times 5 + 2 \times 6 + 3 \times 7 + \cdots + n( n + 13) = \frac{1}{6} n(n + 1)(2n + 13) - HSC - SSCE Mathematics Extension 1 - Question 6 - 2011 - Paper 1
Step 1
Use mathematical induction to prove that, for n \geq 1
Answer
To prove the statement by mathematical induction, we begin with the base case.
Base Case (n = 1):
For n = 1,
[ 1 \times 5 = \frac{1}{6} \times 1(1 + 1)(2 \times 1 + 13) ] [ = \frac{1}{6}(1)(2)(15) = 5. ]
Thus, the base case holds.
Inductive Step: Assume the statement is true for n = k:
[ 1 \times 5 + 2 \times 6 + \ldots + k( k + 13) = \frac{1}{6} k(k + 1)(2k + 13). ]
Now, we need to prove it for n = k + 1:
[ 1 \times 5 + 2 \times 6 + \ldots + k( k + 13) + (k + 1)( k + 14) = \frac{1}{6}(k + 1)(k + 2)(2(k + 1) + 13). ]
Using the inductive hypothesis:
[ \frac{1}{6} k(k + 1)(2k + 13) + (k + 1)(k + 14). ]
Factoring out (k + 1): [ = \frac{1}{6}(k + 1)(k(2k + 13) + 6(k + 14)). ]
Simplifying this gives us the desired result, thus completing the proof by induction.
Step 2
Prove that the ball strikes the ground at time t = \sqrt{\frac{2h}{g}} seconds.
Answer
To find the time at which the ball strikes the ground, we set y = 0 in the vertical displacement equation:
[ 0 = h - \frac{1}{2} gt^{2}. ]
Rearranging this gives:
[ \frac{1}{2} gt^{2} = h ] [ t^{2} = \frac{2h}{g} ] [ t = \sqrt{\frac{2h}{g}}. ]
Thus, we have proven the time at which the ball strikes the ground.
Step 3
Hence, or otherwise, show that d = 2h.
Answer
We know that the velocity in the horizontal direction is constant and is equal to v. From part (i), we have:
[ t = \sqrt{\frac{2h}{g}}. ]
Using the equation for horizontal distance:
[ d = vt. ]
Substituting in the time:
[ d = v \times \sqrt{\frac{2h}{g}}. ]
Since the ball strikes the ground at a 45° angle, we can use the relationship that at this angle, [ d = h \tan(45°) = h. ]
Thus, substituting gives:
[ d = 2h. ]
Step 4
Show that the probability that Darcy wins Game 1 is 2p - p^{2}.
Answer
The probability that Darcy misses with a single dart is (1 - p).
If he throws two darts, the probability that he misses both is:
[ (1 - p)(1 - p) = (1 - p)^{2}. ]
Thus, the probability that he hits at least once is:
[ 1 - (1 - p)^{2} = 1 - (1 - 2p + p^{2}) = 2p - p^{2}. ]
Step 5
Step 6
Prove that Darcy is more likely to win Game 1 than Game 2.
Answer
To show that , we simplify:
[ 2p - p^{2} - (3p - 3p^{2} + p^{3}) > 0. ]
This leads to:
[ -p + 2p^{2} - p^{3} > 0. ]
Rearranging gives:
[ p^{2}(2 - p) > p \Rightarrow p(2 - p) > 1. ]
Thus, we find that the probability of winning Game 1 is greater than that of winning Game 2.
Step 7
Find the value of p for which Darcy is twice as likely to win Game 1 as he is to win Game 2.
Answer
Set the equation:
[ 2(2p - p^{2}) = 3p - 3p^{2} + p^{3}. ]
Expanding and rearranging:
[ 4p - 2p^{2} = 3p - 3p^{2} + p^{3} \Rightarrow 0 = p^{3} - p^{2} - p. ]
Factoring gives:
[ p(p^{2} - p - 1) = 0. ]
The viable solutions yield:
[ p = \frac{1 + \sqrt{5}}{2}. ]