Binomial Theorem Questions and Answers

Answer: E

The coefficient of \(x^{5}\) is 1.

The general term of \(\left(\sqrt{x} + \frac{1}{x}\right)^{10}\) is

\[ \begin{aligned} T_{r+1} &= \binom{10}{r} (x^{1/2})^{10-r} (x^{-1})^{r} \\ &= \binom{10}{r} x^{(10 - 3r)/2} \end{aligned} \]

For \(x^{5}\), \(\frac{10 - 3r}{2} = 5\), so r = 0. The term is the very first one, \(\binom{10}{0}(\sqrt{x})^{10} = x^{5}\), whose coefficient is 1.

The other numbers are coefficients of other terms: 10 and 45 are \(\binom{10}{1}\) and \(\binom{10}{2}\), the coefficients of \(x^{7/2}\) and \(x^{2}\).

Answer: B

The sum is \(2^{15}\).

The sum of all binomial coefficients of order 15 is \(2^{15}\):

\[ \begin{aligned} &{}^{15}C_{0} + {}^{15}C_{1} + \cdots + {}^{15}C_{15} \\ &= 2^{15} \end{aligned} \]

The given expression is missing only \({}^{15}C_{0}\) and \({}^{15}C_{15}\), each equal to 1, and the leading 2 supplies exactly these. So it equals the full sum, \(2^{15}\). \(2^{14}\) would be the sum of the even-index (or odd-index) coefficients alone, and \(2^{16}\) belongs to order 16.

Answer: A

The sum is 31.

Let \(P(x) = (1 + x - 2x^{2})^{6}\). Adding \(P(1)\) and \(P(-1)\) cancels the odd coefficients and doubles the even ones (including the constant term 1):

\[ \begin{aligned} P(1) &= (1 + 1 - 2)^{6} = 0 \\ P(-1) &= (1 - 1 - 2)^{6} = 64 \\ 1 + a_{2} + a_{4} + \cdots + a_{12} &= \frac{0 + 64}{2} \\ &= 32 \\ a_{2} + a_{4} + \cdots + a_{12} &= 31 \end{aligned} \]

The trap is option 32, which forgets to remove the constant term \(a_{0} = 1\). 63 and 64 come from taking \(P(-1)\) without halving.

Answer: E

The sum is \(-510\).

Put \(x = 1\) to get the sum of all coefficients, then remove \(a_{0}\) and \(a_{9}\).

\[ \begin{aligned} a_{0} + a_{1} + \cdots + a_{9} &= (2 - 1)^{9} = 1 \\ a_{0} &= 2^{9} = 512 \\ a_{9} &= (-1)^{9} = -1 \end{aligned} \]

\[ \begin{aligned} a_{1} + \cdots + a_{8} &= 1 - 512 - (-1) \\ &= -510 \end{aligned} \]

The option \(-511\) forgets to remove \(a_{9}\), and \(-512\) subtracts \(a_{9}\) with the wrong sign.

Answer: C

The coefficient is \(-160\).

First find \(n\). The \(x^{3}\) term of \((2 + x)^{n}\) is \(\binom{n}{3}2^{n-3}x^{3}\).

\[ \binom{n}{3}2^{n-3} = 160 \Rightarrow n = 6 \]

since \(\binom{6}{3} \times 2^{3} = 20 \times 8 = 160\). In \((2 - x^{2})^{6}\), \(x^{6}\) comes from the term with \((-x^{2})^{3}\):

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

The option 160 drops the minus sign from \((-1)^{3}\).

Answer: C

a = \(-\frac{1}{2}\).

The general term of \((4x + a)^{7}\) is \(\binom{7}{r}(4x)^{7 - r}a^{r}\). The \(x^{4}\) term has 7 − r = 4, so r = 3.

\[ \begin{aligned} \binom{7}{3}4^{4}a^{3} &= -1120 \\ 35 \times 256\,a^{3} &= -1120 \\ a^{3} &= -\frac{1}{8} \\ a &= -\frac{1}{2} \end{aligned} \]

The negative coefficient needs a negative a, which rules out the positive options; \(-\frac{1}{8}\) is \(a^{3}\), not a.

Answer: E

The coefficient is \(\frac{-4}{9}\).

The general term of \(\left(3x - \frac{1}{3x}\right)^{4}\) is

\[ \begin{aligned} T_{k+1} &= {}^{4}C_{k}(3x)^{4-k}\left(-\tfrac{1}{3x}\right)^{k} \\ &= {}^{4}C_{k}(-1)^{k}\,3^{4-2k}\,x^{4-2k} \end{aligned} \]

For \(x^{-2}\) we need \(4 - 2k = -2\), so \(k = 3\):

\[ {}^{4}C_{3}(-1)^{3}3^{-2} = -\frac{4}{9} \]

The option \(\frac{4}{9}\) drops the sign from \((-1)^{3}\); the others use the wrong term or power of 3.

Answer: A

The coefficient of \(x^{8}\) is -20.

With \(a = x^{2}\) and \(b = \sqrt{1 - x^{2}}\), the odd powers of \(b\) cancel in \((a + b)^{5} + (a - b)^{5}\):

\[ \begin{aligned} (a+b)^{5} + (a-b)^{5} &= 2(a^{5} + 10a^{3}b^{2} + 5ab^{4}) \\ &= 2[x^{10} + 10x^{6}(1 - x^{2}) \\ &\quad + 5x^{2}(1 - x^{2})^{2}] \end{aligned} \]

Only \(10x^{6}(1 - x^{2})\) produces \(x^{8}\), with coefficient \(-10\); the last term reaches only \(x^{6}\). Doubling gives \(-20\). Forgetting the factor 2 gives \(-10\).

Answer: A

\(x\) is 1.

The general term of \((2 + x)^{50}\) is \(T_{r+1} = \binom{50}{r}2^{50-r}x^{r}\). So

\[ \begin{aligned} T_{17} &= \binom{50}{16}2^{34}x^{16} \\ T_{18} &= \binom{50}{17}2^{33}x^{17} \end{aligned} \]

Setting them equal and cancelling \(2^{33}x^{16}\):

\[ \begin{aligned} 2\binom{50}{16} &= \binom{50}{17}x \\ x &= 2 \cdot \frac{17}{34} = 1 \end{aligned} \]

using \(\binom{50}{17} / \binom{50}{16} = \frac{34}{17} = 2\). Taking \(T_{17}\) as the term with \(x^{17}\) (an off-by-one slip) leads to a different value.

Answer: A

The value is \(\alpha = \frac{1}{2}\).

The general term of \((2\alpha - \frac{y}{2})^{8}\) is \(\binom{8}{r}(2\alpha)^{8-r}\left(-\frac{y}{2}\right)^{r}\). For \(y^{3}\), r = 3.

\[ \begin{aligned} T_4 &= \tbinom{8}{3}(2\alpha)^{5}\left(-\tfrac{1}{2}\right)^{3}y^{3} \\ &= 56 \times 32\alpha^{5} \times \left(-\tfrac{1}{8}\right)y^{3} \\ &= -224\alpha^{5}y^{3} \end{aligned} \]

Setting \(-224\alpha^{5} = -7\) gives \(\alpha^{5} = \frac{1}{32}\), so \(\alpha = \frac{1}{2}\). Dropping the minus sign from \(\left(-\tfrac{1}{2}\right)^{3}\) would make \(\alpha\) negative, and the whole-number options make the coefficient far too large (\(\alpha = 2\) gives -7168).