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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

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When expanded, which expression has a non-zero constant term? A. $(x + \frac{1}{x^2})^{7}$ B. $(x^2 + \frac{1}{x^3})^{7}$ C. $(x^3 + \frac{1}{x^4})^{7}$ D. $(x^4... 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. $(x + \frac{1}{x^2})^{7}$

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Answer

To find the constant term in this expansion, we can use the binomial theorem. The general term in the expansion is given by:

Tk=(7k)x7k(1x2)k=(7k)x73kx2kT_k = \binom{7}{k} x^{7-k} \left( \frac{1}{x^2} \right)^{k} = \binom{7}{k} \frac{x^{7 - 3k}}{x^{2k}}

For the term to be constant, the exponent of xx must equal zero:

73k=0k=737 - 3k = 0 \Rightarrow k = \frac{7}{3}

Since kk is not an integer, this expression does not have a constant term.

Step 2

B. $(x^2 + \frac{1}{x^3})^{7}$

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Answer

Using the binomial theorem, the general term is:

Tk=(7k)(x2)7k(1x3)k=(7k)x145kT_k = \binom{7}{k} (x^2)^{7-k} \left( \frac{1}{x^3} \right)^{k} = \binom{7}{k} x^{14 - 5k}

Setting the exponent of xx to zero:

145k=0k=14514 - 5k = 0 \Rightarrow k = \frac{14}{5}

Again, kk is not an integer, so there is no constant term.

Step 3

C. $(x^3 + \frac{1}{x^4})^{7}$

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Answer

The general term is:

Tk=(7k)(x3)7k(1x4)k=(7k)x217kT_k = \binom{7}{k} (x^3)^{7-k} \left( \frac{1}{x^4} \right)^{k} = \binom{7}{k} x^{21 - 7k}

Setting the exponent of xx to zero:

217k=0k=321 - 7k = 0 \Rightarrow k = 3

Since kk is an integer, this expression has a constant term.

Step 4

D. $(x^4 + \frac{1}{x^5})^{7}$

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Answer

The general term is:

Tk=(7k)(x4)7k(1x5)k=(7k)x289kT_k = \binom{7}{k} (x^4)^{7-k} \left( \frac{1}{x^5} \right)^{k} = \binom{7}{k} x^{28 - 9k}

Setting the exponent of xx to zero:

289k=0k=28928 - 9k = 0 \Rightarrow k = \frac{28}{9}

Once more, kk is not an integer, so this expression does not have a constant term.

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