When expanded, which expression has a non-zero constant term? A - HSC - SSCE Mathematics Extension 1 - Question 9 - 2017 - Paper 1

Using the same method, we identify the general term:

$T_k = \binom{7}{k} (x^2)^{(7-k)} \left(\frac{1}{x^3}\right)^k$

The exponent of $x$ simplifies to:

$2(7-k) - 3k = 14 - 5k$

Setting this equal to zero:

$14 - 5k = 0 \\ k = \frac{14}{5}$

Thus, there is also no constant term for this expression.

The general term is:

$T_k = \binom{7}{k} (x^3)^{(7-k)} \left(\frac{1}{x^4}\right)^k$

The exponent of $x$ becomes:

$3(7-k) - 4k = 21 - 7k$

For a constant term, we set:

$21 - 7k = 0 \\ k = 3$

Substituting $k = 3$, we find a constant term exists.

For this expression, the general term is:

$T_k = \binom{7}{k} (x^4)^{(7-k)} \left(\frac{1}{x^5}\right)^k$

The exponent simplifies to:

$4(7-k) - 5k = 28 - 9k$

Setting this equal to zero gives:

$28 - 9k = 0 \\ k = \frac{28}{9}$

Hence, there is no constant term here either.