Binomial Theorem Questions and Answers

Answer: E

\(\alpha - \beta = \) 0.

\(\alpha\) is the sum of the even-indexed coefficients \({}^{10}C_{0}, {}^{10}C_{2}, \ldots, {}^{10}C_{10}\), and \(\beta\) the odd-indexed ones \({}^{10}C_{1}, \ldots, {}^{10}C_{9}\). Put \(x = -1\) in \((1 + x)^{10}\):

\[ \begin{aligned} (1 - 1)^{10} &= \sum_{r=0}^{10} (-1)^{r}\,{}^{10}C_{r} \\ 0 &= \alpha - \beta \end{aligned} \]

Each of \(\alpha\) and \(\beta\) separately equals \(2^{9} = 512\), half of \(2^{10}\). Options 32, 64, 128 and 256 confuse the difference with one of these sums.

Answer: C

\(\alpha = \) 4.

The general term of \((2x + \alpha)^{8}\) is \({}^{8}C_{r}(2x)^{r}\alpha^{8 - r}\). So

\[ \begin{aligned} \text{coeff of } x^{2} &= {}^{8}C_{2}\cdot 4 \cdot \alpha^{6} = 112\alpha^{6} \\ \text{coeff of } x^{3} &= {}^{8}C_{3}\cdot 8 \cdot \alpha^{5} = 448\alpha^{5} \end{aligned} \]

Setting them equal (with \(\alpha \neq 0\)) gives \(112\alpha = 448\), so \(\alpha = 4\). Forgetting the powers of 2 from \((2x)^{r}\) gives \(28\alpha = 56\), \(\alpha = 2\); inverting the ratio gives 1/4.

Answer: E

The constant term is 6000.

The general term of \(\left(2x - \frac{5}{x^{2}}\right)^{6}\) is

\[ \begin{aligned} T_{r+1} \\ &= {}^{6}C_{r}(2x)^{6-r}\left(-\frac{5}{x^{2}}\right)^{r} \end{aligned} \]

with power of \(x\) equal to \(6 - 3r\). This is zero when \(r = 2\):

\[ T_{3} = 15 \times 2^{4} \times 25 = 6000 \]

The sign is positive because \((-5)^{2}\) is positive. The other options come from using the wrong \(r\) or dropping a factor.

Answer: B

The second sum is 256.

The binomial coefficients of a power add to a power of 2: \(\sum_{k=0}^{m} {}^{m}C_{k} = 2^{m}\). So

\[ \begin{aligned} 2^{n+1} &= 512 = 2^{9} \;\Rightarrow\; n = 8 \\ \sum_{k=0}^{8} {}^{8}C_{k} &= 2^{8} = 256 \end{aligned} \]

Each step down in \(n\) halves the sum. 512 repeats the given value, 128 halves it twice, and 511 or 510 are not powers of 2.

Here \(n = 8\) itself is not offered as an option; the question asks for the sum, \(2^{8}\).

Answer: D

The sum is \(2^{100}\).

The first 1 is \({}^{100}C_{0}\) and the last 1 is \({}^{100}C_{100}\), so the expression is the full sum of binomial coefficients of order 100.

\[ \sum_{r=0}^{100} {}^{100}C_{r} = (1 + 1)^{100} = 2^{100} \]

\(2^{99}\) is half that sum (the even or the odd coefficients alone), and \(2^{101}\) doubles it. \(100^{2}\) has no connection with the binomial sum.

Answer: E

The coefficient is 0.

Every term of \((1 - x^{3})^{9}\) has a power of x that is a multiple of 3: 0, 3, 6, 9, ... Multiplying by \(1 - x^{2}\) gives powers of the form 3k (from the 1) or 3k + 2 (from \(-x^{2}\)).

To get \(x^{10}\) we would need 3k = 10 or 3k + 2 = 10, that is k = 10/3 or k = 8/3. Neither is a whole number, so no term contains \(x^{10}\). The binomial coefficients in the other options belong to powers such as \(x^{12}\) or \(x^{14}\), not to \(x^{10}\).

Answer: E

The sum is \(2^{20} - 1\).

The 22 coefficients of order 21 add to \(2^{21}\), and they pair up by symmetry: \({}^{21}C_{r} = {}^{21}C_{21-r}\). So \({}^{21}C_{0}\) to \({}^{21}C_{10}\) make up exactly half.

\[ \begin{aligned} \sum_{r=0}^{10} {}^{21}C_{r} &= \frac{2^{21}}{2} = 2^{20} \\ \sum_{r=1}^{10} {}^{21}C_{r} &= 2^{20} - 1 \end{aligned} \]

The sum starts at r = 1, so the term \({}^{21}C_{0} = 1\) is removed. \(2^{20}\) forgets that, and \(2^{21} - 1\) or \(2^{21} - 2\) treat the sum as if it ran over all the coefficients.

Answer: C

The constant term is \(\frac{5}{256}\).

The general term is

\[ \begin{aligned} T_{r+1} &= {}^{10}C_{r}\left(\frac{\sqrt{x}}{2}\right)^{10-r} \\ &\quad \times \left(\frac{1}{3x^{2}}\right)^{r} \end{aligned} \]

with power of x equal to \(\frac{10 - r}{2} - 2r\).

\[ \begin{aligned} \frac{10 - r}{2} - 2r = 0 &\Rightarrow r = 2 \\ T_{3} = {}^{10}C_{2}\cdot\frac{1}{2^{8}}\cdot\frac{1}{3^{2}} &= \frac{45}{256 \times 9} = \frac{5}{256} \end{aligned} \]

Using \(2^{7} = 128\) instead of \(2^{8}\) gives 5/128, and forgetting to square the 3 leaves a 9 or 15 in the wrong place. The term exists, so 0 is wrong.

Answer: E

The coefficient of \(x^{9}\) is 0.

Every term of \(\left(4 - \frac{x^{2}}{4}\right)^{12}\) has the form \({}^{12}C_{r}\, 4^{12-r}\left(-\frac{x^{2}}{4}\right)^{r}\), which contains \(x^{2r}\). Only even powers of x appear, so there is no \(x^{9}\) term and its coefficient is 0.

The options built from \({}^{12}C_{7}\), \({}^{12}C_{6}\) and \({}^{12}C_{5}\) would need \(2r = 9\), which has no whole-number solution.

The general term contains \(x^{2r}\), so this binomial has only even powers of x; any question about an odd power of x in it has the answer 0.

Answer: A

\(n = 18\).

The general term of \(\left(x^{2} + \frac{1}{x^{3}}\right)^{n}\) is \({}^{n}C_{r}\, x^{2(n-r)}x^{-3r} = {}^{n}C_{r}\, x^{2n - 5r}\). For the term in x, \(2n - 5r = 1\).

The coefficient is \({}^{n}C_{7}\), so \(r = 7\) (or \(r = n - 7\)). With \(r = 7\): \(2n = 36\), so \(n = 18\). (The alternative \(r = n - 7\) gives \(2n - 5n + 35 = 1\), \(n = \frac{34}{3}\), not an integer.)

Check: for \(n = 18\), \(r = 7\) gives the power \(36 - 35 = 1\). The other values of n leave no integer r with coefficient \({}^{n}C_{7}\).