Binomial Theorem Questions and Answers

Answer: E

The value of a is 2.

Expand \((1 - 2x)^{3} = 1 - 6x + 12x^{2} - 8x^{3}\). In the product with \((1 + ax)\), the \(x^{2}\) terms come from \(1 \times 12x^{2}\) and \(ax \times (-6x)\).

\[ \begin{aligned} a_2 &= 12 - 6a = 0 \\ a &= 2 \end{aligned} \]

Taking the \(x^{2}\) coefficient as 12 alone, or pairing ax with the \(x^{3}\) term, ignores how the product forms \(x^{2}\); only \(12 - 6a = 0\) uses both contributions.

Answer: A

The constant term is −160.

The general term of \(\left(2x^{2} - \frac{1}{x^{2}}\right)^{6}\) is

\[ \begin{aligned} T_{r+1} &= {}^{6}C_{r}(2x^{2})^{6-r} \\ &\quad \times \left(-\frac{1}{x^{2}}\right)^{r} \end{aligned} \]

with power of x \(12 - 4r\). Setting it to 0 gives \(r = 3\):

\[ \begin{aligned} {}^{6}C_{3} \cdot 2^{3} \cdot (-1)^{3} &= 20 \times 8 \times (-1) \\ &= -160 \end{aligned} \]

160 drops the sign from \((-1)^{3}\), and ±180 use the wrong binomial coefficient or power of 2.

Answer: D

The coefficient of x³ is 540.

The general term of \((x^{5/2} + 3x^{-3/2})^{6}\) is:

\[ \begin{aligned} T_{r+1} &= \binom{6}{r}(x^{5/2})^{6-r}(3x^{-3/2})^{r} \\ &= \binom{6}{r}3^{r}x^{15 - 4r} \end{aligned} \]

For x³, 15 - 4r = 3, so r = 3, and the coefficient is \(\binom{6}{3} \times 3^{3} = 20 \times 27 = 540\). 120 forgets the 3³ (or uses a wrong binomial), and 0 would mean no term has power 3, which is not the case.