Algebra - Binomial Theorem

Find $P^n_2$ if the fifth term of the expansion of $\left(\sqrt[3]{x} + \dfrac{1}{x}\right)^n$ does not depend on $x$.

Fifth term of the expansion of $\left(\sqrt[3]{x} + \dfrac{1}{x}\right)^n$ is

$T_5 = T_{4+1} = C^n_4 \; \left(x^{\frac{1}{3}}\right)^{n-4} \; \left(x^{-1}\right)^4$

i.e. $\;$ $T_5 = C^n_4 \; x^{\left(\frac{n}{3} - \frac{4}{3} - 4\right)}$

i.e. $\;$ $T_5 = C^n_4 \; x^{\left(\frac{n-16}{3}\right)}$

As per question, the fifth term in the binomial expansion is independent of $x$.

$\implies$ $\dfrac{n-16}{3} = 0$ $\implies$ $n = 16$

Now, $\;$ $P^n_2 = P^{16}_2 = \dfrac{16!}{\left(16 - 2\right)!} = \dfrac{16 \times 15 \times 14!}{14!} = 240$

Algebra - Binomial Theorem

The sum of the coefficients in the first three terms of the expansion of $\left(x^2 - \dfrac{2}{x}\right)^m$ is equal to $97$. Find the term of the expansion containing $x^4$.

In the expansion of $\left(x^2 - \dfrac{2}{x}\right)^m$,

first term $= T_1 = C^m_0 \; \left(x^2\right)^m = C^m_0 \times x^{2m}$

second term $= T_2 = C^m_1 \; \left(x^2\right)^{m-1} \; \left(\dfrac{-2}{x}\right) = -2 C^m_1 \times x^{2m-3}$

third term $= T_3 = C^m_2 \; \left(x^2\right)^{m-2} \; \left(\dfrac{-2}{x}\right)^2 = 4 \; C^m_2 \times x^{2m-6}$

$\therefore \;$ Coefficients in the first three terms of the expansion of $\left(x^2 - \dfrac{2}{x}\right)^m$ are $\;$ $C^m_0, \; -2 C^m_1, \; 4C^m_2$.

As per question, $\;\;\;$ $C^m_0 - 2 C^m_1 + 4 C^m_2 = 97$

i.e. $\;$ $1 - \dfrac{2 \times m!}{1! \left(m-1\right)!} + \dfrac{4 \times m!}{2! \left(m-2\right)!} = 97$

i.e. $\;$ $-2m + \dfrac{4m \left(m-1\right)}{2} = 96$

i.e. $\;$ $2m^2 - 4m = 96$

i.e. $\;$ $m^2 - 2m - 48 = 0$

i.e. $\;$ $\left(m-8\right) \left(m+6\right) = 0$

i.e. $\;$ $m = 8$ $\;$ or $\;$ $m = -6$

$\because \;$ $m$ cannot be negative $\implies$ $m = 8$

Let $\left(r+1\right)^{th}$ term in the expansion of $\left(x^2 - \dfrac{2}{x}\right)^m = \left(x^2 - \dfrac{2}{x}\right)^8$ $\;$ be the term of the expansion containing $x^4$.

Now, $\;$ $\left(r+1\right)^{th}$ term $= T_{r+1} = C^8_r \; \left(x^2\right)^{8-r} \; \left(\dfrac{-2}{x}\right)^r$

i.e. $\;$ $T_{r+1} = \left(-2\right)^r \; C^8_r \; x^{16-3r}$

$\therefore \;$ As per question, $\;$ $16 - 3r = 4$

i.e. $\;$ $3r = 12$ $\implies$ $r = 4$

$\therefore \;$ The term containing $x^4$ in the expansion of $\left(x^2 - \dfrac{2}{x}\right)^8$ is

$T_{4+1} = T_5 = \left(-2\right)^4 \; C^8_4 \; x^4$

i.e. $\;$ $T_5 = 16 \times \dfrac{8!}{4! \left(8-4\right)!} \times x^4$

i.e. $\;$ $T_5 = \dfrac{16 \times 8 \times 7 \times 6 \times 5 \times x^4}{4 \times 3 \times 2 \times 1} = 1120 x^4$

Algebra - Binomial Theorem

Find the term of the expansion of $\left(\sqrt[3]{x} - \dfrac{1}{\sqrt{x}}\right)^{15}$ which does not contain $x$.

$\left(k + 1\right)^{th}$ term in the expansion of $\left(a + b\right)^n$ is \;\; $T_{k+1} = C^n_k \; a^{n-k} \; b^k$

Let the $\left(k + 1\right)^{th}$ term in the expansion of $\left(\sqrt[3]{x} - \dfrac{1}{\sqrt{x}}\right)^{15} = \left(x^{\frac{1}{3}} - x^{\frac{-1}{2}}\right)^{15}$ be independent of $x$.

Now, $\;$ $T_{k+1} = C^{15}_k \; \left(x^{\frac{1}{3}}\right)^{15 - k} \; \left(x^{\frac{-1}{2}}\right)^{k} = C^{15}_k \; x^{5- \frac{k}{3} - \frac{k}{2}}$

Since this term is independent of $x$ $\implies$ $5 - \dfrac{k}{3} - \dfrac{k}{2} = 0$

i.e. $\;$ $5 = \dfrac{5k}{6}$ $\implies$ $k = 6$

i.e. $7^{th}$ term is independent of $x$.

$\therefore \;$ The term independent of $x$ in the expansion of $\left(\sqrt[3]{x} - \dfrac{1}{\sqrt{x}}\right)^{15}$ is

$T_{7} = T_{6+1} = C^{15}_{6} = \dfrac{15!}{6! \left(15 - 6\right)!} = \dfrac{15!}{6! \times 9!} = 5005$

Algebra - Binomial Theorem

Determine the ordinal number of the term of the expansion of $\left(\dfrac{3}{4} \sqrt[3]{a^2} + \dfrac{2}{3} \sqrt{a}\right)^{12}$ which contains $a^7$.

In the expansion of $\left(x+y\right)^n$, $\;$ $\left(r+1\right)^{th}$ term $= T_{r+1} = C^n_r \; x^{n-r} \; y^r$

$\left(r+1\right)^{th}$ term in the expansion of $\left(\dfrac{3}{4} \sqrt[3]{a^2} + \dfrac{2}{3} \sqrt{a}\right)^{12} = \left(\dfrac{3}{4} a^{\frac{2}{3}} + \dfrac{2}{3} a^{\frac{1}{2}}\right)^{12}$ is

$T_{r+1} = C^{12}_r \; \left(\dfrac{3}{4} a^{\frac{2}{3}}\right)^{12-r} \; \left(\dfrac{2}{3} a^{\frac{1}{2}}\right)^r$

i.e. $\;$ $T_{r+1} = C^{12}_r \times \dfrac{3^{12-r} \; a^{\frac{24-2r}{3}}}{2^{24-2r}} \times \dfrac{2^r \; a^{\frac{r}{2}}}{3^r}$

i.e. $\;$ $T_{r+1} = C^{12}_r \times \dfrac{3^{12-2r} \; a^{\frac{24-2r}{3} + \frac{r}{2}}}{2^{24-3r}}$

Let the $\left(r+1\right)^{th}$ term in the expansion contain $a^7$.

Then,

$\dfrac{24-2r}{3} + \dfrac{r}{2} = 7$

i.e. $\;$ $48 - 4r + 3r = 42$

i.e. $\;$ $r = 6$

$\therefore \;$ The ordinal number of the term in the expansion of $\left(\dfrac{3}{4} \sqrt[3]{a^2} + \dfrac{2}{3} \sqrt{a}\right)^{12}$ which contains $a^7$ is $6$.

Algebra - Binomial Theorem

Find $x$ in the binomial expansion $\left(\sqrt[3]{2} + \dfrac{1}{\sqrt[3]{3}}\right)^x$ if the ratio of the seventh term from the beginning of the binomial expansion to the seventh term from its end is $\dfrac{1}{6}$.

In the expansion of $\left(x+a\right)^n$

$\left(r+1\right)^{th}$ term from the beginning $\;$ $= T_{r+1} = C^n_r \; x^{n-r} \; a^r$

$r^{th}$ term from the end in the expansion of $\left(x+a\right)^n$ is the $\left(n - r + 2\right)^{th}$ term from the beginning

i.e. $\;$ $r^{th}$ term from the end $\;$ $= T_{n-r+2}$

Now, in the expansion of $\left(\sqrt[3]{2} + \dfrac{1}{\sqrt[3]{3}}\right)^x$

$7^{th}$ term from the beginning $= T_7 = T_{6+1} = C^x_6 \; \left(\sqrt[3]{2}\right)^{x-6} \; \left(\dfrac{1}{\sqrt[3]{3}}\right)^6$

$7^{th}$ term from the end $= T_{x-7+2} = T_{x-5} = T_{x-6+1} = C^x_{x-6} \; \left(\sqrt[3]{2}\right)^{x-x+6} \; \left(\dfrac{1}{\sqrt[3]{3}}\right)^{x-6}$

i.e. $\;$ $T_{x-5} = C^x_{x-6} \; \left(\sqrt[3]{2}\right)^6 \; \left(\dfrac{1}{\sqrt[3]{3}}\right)^{x-6}$

As per question,

$\dfrac{T_7}{T_{x-5}} = \dfrac{1}{6}$

i.e. $\;$ $\dfrac{C^x_6 \; \left(\sqrt[3]{2}\right)^{x-6} \; \left(\dfrac{1}{\sqrt[3]{3}}\right)^6}{C^x_{x-6} \; \left(\sqrt[3]{2}\right)^6 \; \left(\dfrac{1}{\sqrt[3]{3}}\right)^{x-6}} = \dfrac{1}{6}$

Since $\;$ $C^n_r = C^n_{n-r}$ $\implies$ $C^x_6 = c^x_{x-6}$

$\therefore \;$ We have

$\left(\sqrt[3]{2}\right)^{x-12} \times \left(\dfrac{1}{\sqrt[3]{3}}\right)^{12-x} = \dfrac{1}{6}$

i.e. $\;$ $\dfrac{2^{\frac{x}{3}} \times 2^{\frac{-12}{3}}}{3^{\frac{12}{3}} \times 3^{\frac{-x}{3}}} = \dfrac{1}{6}$

i.e. $\;$ $\dfrac{2^{\frac{x}{3}} \times 2^{-4}}{3^{4} \times 3^{\frac{-x}{3}}} = \dfrac{1}{6}$

i.e. $\;$ $\dfrac{2^{\frac{x}{3}} \times 3^{\frac{x}{3}}}{2^4 \times 3^4} = \dfrac{1}{2 \times 3}$

i.e. $\;$ $2^{\frac{x}{3}} \times 3^{\frac{x}{3}} = 2^3 \times 3^3$

$\implies$ $\dfrac{x}{3} = 3$

$\implies$ $x = 9$

Algebra - Binomial Theorem

Find the third term of the expansion of $\left(z^2 + \dfrac{1}{z} \sqrt[3]{z}\right)^n$ if the sum of all the binomial coefficients is equal to $2048$.

Sum of all binomial coefficients is

$C^n_0 + C^n_1 + C^n_2 + \cdots + C^n_n = 2^n$

Then, as per question,

$2^n = 2048$

i.e. $\;$ $2^n = 2^{11}$ $\implies$ $n = 11$

$\therefore \;$ the problem is to find the third term in the expansion of

$\begin{aligned} \left(z^2 + \dfrac{1}{z} \sqrt[3]{z}\right)^n & = \left(z^2 + z^{\frac{1}{3} - 1}\right)^{11} \\\\ & = \left(z^2 + z^{\frac{-2}{3}}\right)^{11} \end{aligned}$

$\therefore \;$ The required third term is

$\begin{aligned} T_3 = T_{2+1} & = C^{11}_{2} \; \left(z^2\right)^{11 - 2} \; \left(z^{\frac{-2}{3}}\right)^2 \\\\ & = \dfrac{11!}{2! \times \left(11 - 2\right)!} \times \left(z^2\right)^9 \times z^{\frac{-4}{3}} \\\\ & = \dfrac{11 \times 10 \times 9!}{2 \times 9!} \times z^{18 - \frac{4}{3}} \\\\ & = 55 z^{\frac{50}{3}} \end{aligned}$

Algebra - Binomial Theorem

Find the second term of the binomial expansion of $\left(\sqrt[13]{a} + \dfrac{a}{\sqrt{a^{-1}}}\right)^m$ if $C^m_3 : C^m_2 = 4 : 1$.

Given: $\;\;$ $C^m_3 : C^m_2 = 4 : 1$

i.e. $\;$ $\dfrac{m!}{3! \left(m-3\right)!} : \dfrac{m!}{2! \left(m-2\right)!} = \dfrac{4}{1}$

i.e. $\;$ $\dfrac{m!}{3 \times 2! \left(m-3\right)!} \times \dfrac{2! \left(m-2\right) \left(m-3\right)!}{m!} = 4$

i.e. $\;$ $\dfrac{m-2}{3} = 4$

i.e. $\;$ $m - 2 = 12$ $\implies$ $m = 14$

$\therefore \;$ The problem is: $\;\;$ $\left(\sqrt[13]{a} + \dfrac{a}{\sqrt{a^{-1}}}\right)^m = \left(a^{\frac{1}{13}} + a^{1 + \frac{1}{2}}\right)^{14} = \left(a^{\frac{1}{13}} + a^{\frac{3}{2}}\right)^{14}$

$\therefore \;$ Second term in the expansion of $\left(a^{\frac{1}{13}} + a^{\frac{3}{2}}\right)^{14}$ is

$\begin{aligned} T_2 = T_{1+1} & = C^{14}_{1} \; \left(a^{\frac{1}{13}}\right)^{14-1} \; \left(a^{\frac{3}{2}}\right)^1 \\\\ & = 14 \times \left(a^{\frac{1}{13}}\right)^{13} \times a^{\frac{3}{2}} \\\\ & = 14 \; a^{1+\frac{3}{2}} \\\\ & = 14a^{\frac{5}{2}} \end{aligned}$

Algebra - Binomial Theorem

Find the term of the expansion of $\left(\sqrt[3]{x^{-2}} + x\right)^7$ containing $x$ in the second power.

$\left(r+1\right)^{th}$ term in the expansion of $\left(a+b\right)^n$ is $\;\;$ $T_{r+1} = C^n_r \; a^{n-r} \; b^r$

Here, $\;$ $n = 7, \; a = \sqrt[3]{x^{-2}} = x^{\frac{-2}{3}}, \; b = x$

Let the $\left(r+1\right)^{th}$ term in the expansion of $\left(\sqrt[3]{x^{-2}} + x\right)^7$ contain $x$ in the second power.

Now, $\left(r+1\right)^{th}$ term is

$\begin{aligned} T_{r+1} & = C^7_r \; \left(x^{\frac{-2}{3}}\right)^{7-r} \; x^r \\\\ & = C^7_r \; x^{\frac{-14}{3} + \frac{2r}{3} + r} \\\\ & = C^7_r \; x^{\frac{5r-14}{3}} \end{aligned}$

$\because \;$ the $\left(r+1\right)^{th}$ term contains $x$ in the second power,

$\implies$ $\dfrac{5r - 14}{3} = 2$

i.e. $\;$ $5r - 14 = 6$

i.e. $\;$ $5r = 20$

$\implies$ $r = 4$

$\therefore \;$ The required term is

$\begin{aligned} T_{4+1} = T_5 & = C^7_4 \; x^{\frac{20 - 14}{3}} \\\\ & = \dfrac{7!}{4! \times 3!} x^2 \\\\ & = \dfrac{7 \times 6 \times 5 \times 4!}{4! \times 6} x^2 \\\\ & = 35 x^2 \end{aligned}$

Algebra - Binomial Theorem

Find the term in the expansion of $\left(x + \dfrac{1}{x}\right)^8$ which does not contain $x$.

By binomial theorem, $\;$ $\left(a + b\right)^n = C^n_0 \; a^n + C^n_1 \; a^{n-1} \; b + \cdots + C^n_k \; a^{n-k} \; b^k + \cdots + C^n_n \; b^n$

$\left(k + 1\right)^{th}$ term in the expansion of $\left(a + b\right)^n$ is \;\; $T_{k+1} = C^n_k \; a^{n-k} \; b^k$

Let the $\left(k + 1\right)^{th}$ term in the expansion of $\left(x + \dfrac{1}{x}\right)^8$ be independent of $x$.

Now, $\;$ $T_{k+1} = C^8_k \; x^{8-k} \; \left(\dfrac{1}{x}\right)^{k} = C^8_k \; x^{8-k-k} = C^8_k \; x^{8-2k}$

Since this term is independent of $x$ $\implies$ $8 - 2k = 0$ $\implies$ $k = 4$

i.e. $5^{th}$ term is independent of $x$.

$\therefore \;$ The term independent of $x$ in the expansion of $\left(x + \dfrac{1}{x}\right)^8$ is

$T_5 = T_{4+1} = C^8_4 = \dfrac{8!}{4! \left(8 - 4\right)!} = \dfrac{8 \times 7 \times 6 \times 5 \times 4!}{4! \times 4 \times 3 \times 2 \times 1} = 70$

Algebra - Binomial Theorem

In the expansion of $\left(a \sqrt{a} + \dfrac{1}{a^4}\right)^n$, the coefficient in the second term exceeds by 44 the coefficient in the first term. Find $n$.

In the expansion of $\left(a \sqrt{a} + \dfrac{1}{a^4}\right)^n$,

coefficient in first term $= C^n_1$; $\;$ coefficient in second term $= C^n_2$

As per question,

$C^n_2 = C^n_1 + 44$

i.e. $\;$ $\dfrac{n!}{2!\left(n-2\right)!} = \dfrac{n!}{\left(n-1\right)!} + 44$

i.e. $\;$ $\dfrac{n \left(n-1\right) \left(n-2\right)!}{2\left(n-2\right)!} = \dfrac{n \left(n-1\right)!}{\left(n-1\right)!} + 44$

i.e. $\;$ $\dfrac{n \left(n-1\right)}{2} = n + 44$

i.e. $\;$ $n \left(\dfrac{n-1}{2} - 1\right) = 44$

i.e. $\;$ $\dfrac{n \left(n-3\right)}{2} = 44$

i.e. $\;$ $n^2 - 3n - 88 = 0$

i.e. $\;$ $\left(n-11\right)\left(n+8\right) = 0$

i.e. $\;$ $n = 11$ $\;$ or $\;$ $n = -8$

$\because \;$ $n$ cannot be negative as it is the total number of objects

$\therefore \;$ $n = 11$

Algebra - Binomial Theorem

Find the middle term of the expansion of $\left(\sqrt{x} - \dfrac{1}{x}\right)^6$

There are $7$ terms in the expansion of $\left(\sqrt{x} - \dfrac{1}{x}\right)^6$.

The $4^{th}$ term will be the middle term.

Now, the $\left(r + 1\right)^{th}$ term $T_{r+1}$ in the expansion of $\left(p + q\right)^n$ is

$T_{r+1} = C^n_r \; p^{n-r} \; q^r$

$\therefore \;$ In the given problem,

Fourth term $= T_4 = T_{3+1} = C^{6}_{3} \; \left(\sqrt{x}\right)^{6-3} \; \left(\dfrac{-1}{x}\right)^3$

$\begin{aligned} i.e. \; T_4 & = \dfrac{6!}{3! \left(6-3\right)!} \times \left(\sqrt{x}\right)^3 \times \left(\dfrac{-1}{x^3}\right) \\\\ & = \dfrac{-6 \times 5 \times 4 \times 3!}{3! \times 6} \times x^{\frac{3}{2} - 3} \\\\ & = -20 x^{\frac{-3}{2}} \end{aligned}$