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When expanded, which expression has a non-zero constant term? A - HSC - SSCE Mathematics Extension 1 - Question 9 - 2017 - Paper 1
Question 9
When expanded, which expression has a non-zero constant term? A. \(\left( x + \frac{1}{x^2} \right)^{7} \) B. \(\left( x^{2} + \frac{1}{x^{3}} \right)^{7} \) C. \... show full transcript
Worked Solution & Example Answer:When expanded, which expression has a non-zero constant term? A - HSC - SSCE Mathematics Extension 1 - Question 9 - 2017 - Paper 1
Step 1
A. \(\left( x + \frac{1}{x^2} \right)^{7} \)
Answer
To find the constant term, we consider the general term in the binomial expansion: ( T_k = \binom{n}{k} x^{n-k} \left(\frac{1}{x^2} \right)^{k} ). Here, when we set the exponent of x to zero: ( n - k - 2k = 0 \Rightarrow n - 3k = 0 ). Solving gives us ( k = \frac{7}{3} ), which is not an integer. Thus, there is no constant term.
Step 2
B. \(\left( x^{2} + \frac{1}{x^{3}} \right)^{7} \)
Answer
Applying the binomial theorem: ( T_k = \binom{n}{k} (x^{2})^{n-k} \left(\frac{1}{x^{3}} \right)^{k} ). We set the exponent of x to zero: ( 2(n-k) - 3k = 0 \Rightarrow 2(7-k) - 3k = 0 \Rightarrow 14 - 5k = 0 \Rightarrow k = \frac{14}{5} ), which is not an integer. So, no constant term.
Step 3
C. \(\left( x^{3} + \frac{1}{x^{4}} \right)^{7} \)
Answer
Using the result from binomial expansion, we have: ( T_k = \binom{7}{k} (x^{3})^{7-k} \left(\frac{1}{x^{4}} \right)^{k} ). We set ( 3(7-k) - 4k = 0 \Rightarrow 21 - 3k - 4k = 0 \Rightarrow 21 - 7k = 0 \Rightarrow k = 3 ), which is an integer. Thus, we have a constant term.
Step 4
D. \(\left( x^{4} + \frac{1}{x^{5}} \right)^{7} \)
Answer
Applying similar analysis, ( T_k = \binom{7}{k} (x^{4})^{7-k} \left(\frac{1}{x^{5}} \right)^{k} ). Setting the power of x to zero: ( 4(7-k) - 5k = 0 \Rightarrow 28 - 4k - 5k = 0 \Rightarrow 28 - 9k = 0 \Rightarrow k = \frac{28}{9} ), which is not an integer. Hence, no constant term.