Find the 4th term from the beginning and end in the expansion

Question: Find the $4^{\text {th }}$ term from the beginning and end in the expansion of $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^{\mathrm{n}}$ Solution: To Find : I. $4^{\text {th }}$ term from the beginning II. $4^{\text {th }}$ term from the end Formulae : $\cdot \mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$ $\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right)=\left(\be...

Find the 4th term from the end in the expansion of

Question: Find the $4^{\text {th }}$ term from the end in the expansion of $\left(\frac{4 x}{5}-\frac{5}{2 x}\right)^{9}$ Solution: To Find : $4^{\text {th }}$ term from the end Formulae : $\cdot \mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$ $\cdot\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right)=\left(\begin{array}{c}\mathrm{n} \\ \mathrm{n}-\mathrm{r}\end{array}\right)$ For $...

Find the 5th term from the end in the expansion of

Question: Find the $5^{\text {th }}$ term from the end in the expansion of $\left(x-\frac{1}{x}\right)^{12}$ Solution: To Find : $5^{\text {th }}$ term from the end Formulae : - $\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$ $\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right)=\left(\begin{array}{c}\mathrm{n} \\ \mathrm{n}-\mathrm{r}\end{array}\right)$ For $\left(x-\frac{1}{x}\ri...

Write the general term in the expansion of

Question: Write the general term in the expansion of $\left(x^{2}-y\right)^{6}$ Solution: To Find : General term, i.e. $t_{r+1}$ For $\left(x^{2}-y\right)^{6}$ $a=x^{2}, b=-y$ and $n=6$ General term $t_{r+1}$ is given by, $\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$ $=\left(\begin{array}{l}6 \\ r\end{array}\right)\left(x^{2}\right)^{6-r}(-y)^{r}$ $\underline{\text { Conclusion : General ter...

Show that the expansion of

Question: Show that the expansion of $\left(x^{2}+\frac{1}{x}\right)^{12}$ does not contain any term involving $x^{-1}$ Solution: For $\left(x^{2}+\frac{1}{x}\right)^{12}$ $\mathrm{a}=\mathrm{x}^{2}, \mathrm{~b}=\frac{1}{\mathrm{x}}$ and $\mathrm{n}=12$ We have a formula, $\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$ $=\left(\begin{array}{c}12 \\ r\end{array}\right)\left(x^{2}\right)^{12-r}\...

Show that the expansion of

Question: Show that the expansion of $\left(2 x^{2}-\frac{1}{x}\right)^{20}$ does not contain any term involving $x^{9}$ Solution: For $\left(2 x^{2}-\frac{1}{x}\right)^{20}$ $a=2 x^{2}, b=\frac{-1}{x}$ and $n=20$ We have a formula, $\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$ $=\left(\begin{array}{c}20 \\ r\end{array}\right)\left(3 x^{2}\right)^{20-r}\left(\frac{-1}{x}\right)^{r}$ $=\left(...

Show that the term containing

Question: Show that the term containing $x^{3}$ does not exist in the expansion of $\left(3 x-\frac{1}{2 x}\right)^{8}$ Solution: For $\left(3 x-\frac{1}{2 x}\right)^{8}$ $a=3 x, b=\frac{-1}{2 x}$ and $n=8$ We have a formula, $t_{r+1}=\left(\begin{array}{l}n \\ r\end{array}\right) a^{n-r} b^{r}$ $=\left(\begin{array}{l}8 \\ r\end{array}\right)(3 x)^{8-r}\left(\frac{-1}{2 x}\right)^{r}$ $=\left(\begin{array}{l}8 \\ r\end{array}\right)(3)^{8-r}(x)^{8-r}\left(\frac{-1}{2}\right)^{r}(x)^{-r}$ $=\lef...

Question: Show that the term containing $x^{3}$ does not exist in the expansion of $\left(3 x-\frac{1}{2 x}\right)^{8}$ Solution: For $\left(3 x-\frac{1}{2 x}\right)^{8}$ $a=3 x, b=\frac{-1}{2 x}$ and $n=8$ We have a formula, $t_{r+1}=\left(\begin{array}{l}n \\ r\end{array}\right) a^{n-r} b^{r}$ $=\left(\begin{array}{l}8 \\ r\end{array}\right)(3 x)^{8-r}\left(\frac{-1}{2 x}\right)^{r}$ $=\left(\begin{array}{l}8 \\ r\end{array}\right)(3)^{8-r}(x)^{8-r}\left(\frac{-1}{2}\right)^{r}(x)^{-r}$ $=\lef...

Find the coefficient of

Question: Find the coefficient of (i) $x^{5}$ in the expansion of $(x+3)^{8}$ (ii) $x^{6}$ in the expansion of $\left(3 x^{2}-\frac{1}{3 x}\right)^{9}$. (iii) $\mathbf{x}^{-15}$ in the expansion of $\left(3 \mathrm{x}^{2}-\frac{\mathrm{a}}{3 \mathrm{x}^{3}}\right)^{10}$. (iv) $a^{7} b^{5}$ in the expansion of $(a-2 b)^{12}$. Solution: (i) Here, $a=x, b=3$ and $n=8$ We have a formula $\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}...

Find the coefficient of x in the expansion of

Question: Find the coefficient of $x$ in the expansion of $\left(1-3 x+7 x^{2}\right)(1-x)^{16}$. Solution: To Find : coefficient of x Formula : $\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$ We have a formula, $\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$ Therefore, expansion of $(1-x)^{...

Find the term independent of x in the expansion of

Question: Find the term independent of x in the expansion of (91 + x +$\left.2 x^{3}\right)$ $\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{9}$ Solution: To Find : term independent of $x$, i.e. coefficient of $x^{0}$ Formula: $\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$ We have a formula $\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{...

Show that the ratio of the coefficient of

Question: Show that the ratio of the coefficient of $x^{10}$ in the expansion of $(1-x 2)^{10}$ and the term independent of x in the expansion of $\left(x-\frac{2}{x}\right)^{10}$ is $1: 32$ Solution: To Prove : coefficient of $x^{10}$ in $\left(1-x^{2}\right)^{10}:$ coefficient of $x^{0}$ in $\left(x-\frac{2}{x}\right)^{10}=1: 32$ For $\left(1-x^{2}\right)^{10}$ Here, $a=1, b=-x^{2}$ and $n=15$ We have formula, $\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}...

Find the ratio of the coefficient of

Question: Find the ratio of the coefficient of $x^{15}$ to the term independent of $x$ in the expansion of $\left(x^{2}+\frac{2}{x}\right)^{15}$ Solution: To Find: the ratio of the coefficient of $x^{15}$ to the term independent of $x$ Formula : $\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$ Here, $a=x^{2}, \quad b=\frac{2}{x}$ and $n=15$ We have a formula, $\mathrm{t}_{\mathrm{r}+1}=\left(\b...

Find the coefficients of

Question: Find the coefficients of $x^{7}$ and $x^{8}$ in the expansion of $\left(2+\frac{x}{3}\right)^{n}$. Solution: To find : coefficients of $x^{7}$ and $x^{8}$ Formula : $\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$ Here, $a=2, b=\frac{x}{3}$ We have, $\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}...

Find the 13th term in the expansion of

Question: Find the $13^{\text {th }}$ term in the expansion of $\left(9 x-\frac{1}{3 \sqrt{x}}\right)^{18}, x \neq 0$ Solution: To find: $13^{\text {th }}$ term in the expansion of $\left(9 x-\frac{1}{3 \sqrt{x}}\right)^{18}$ Formula used: (i) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$ (ii) $T_{r+1}={ }^{n} C_{r} a^{n-r} b^{r}$ For $13^{\text {th }}$ term, $r+1=13$ $\Rightarrow r=12$ $\ln ,\left(9 x-\frac{1}{3 \sqrt{x}}\right)^{18}$ $13^{\text {th }}$ term $=T_{12+1}$ $\Rightarrow{ }^{18} C_{12}(9...

Find the 16th term in the expansion of

Question: Find the $16^{\text {th }}$ term in the expansion of $(\sqrt{x}-\sqrt{y})^{17}$ Solution: To find: $16^{\text {th }}$ term in the expansion of $(\sqrt{x}-\sqrt{y})^{17}$ Formula used: (i) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$ (ii) $T_{r+1}={ }^{n} C_{r} a^{n-r} b^{r}$ For $16^{\text {th }}$ term, $r+1=16$ $\Rightarrow \mathrm{r}=15$ $\ln ,(\sqrt{x}-\sqrt{y})^{17}$ $16^{\text {th }}$ term $=\mathrm{T}_{15+1}$ $\Rightarrow{ }^{17} C_{15}(\sqrt{x})^{17-15}(-\sqrt{y})^{15}$ $\Rightarrow...

Find the 9th term in the expansion of

Question: Find the $9^{\text {th }}$ term in the expansion of $\left(\frac{a}{b}-\frac{b}{2 a^{2}}\right)^{12}$ Solution: To find: $9^{\text {th }}$ term in the expansion of $\left(\frac{a}{b}-\frac{b}{2 a^{2}}\right)^{12}$ Formula used: (i) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$ (ii) $T_{r+1}={ }^{n} C_{r} a^{n-r} b^{r}$ For $9^{\text {th }}$ term, $r+1=9$ $\Rightarrow r=8$ $\ln ,\left(\frac{a}{b}-\frac{b}{2 a^{2}}\right)^{12}$ $9^{\text {th }}$ term $=T_{8+1}$ $\Rightarrow 12 \mathrm{C}_{8}\...

Find the 7th term in the expansion of

Question: Find the $7^{\text {th }}$ term in the expansion of $\left(\frac{4 x}{5}+\frac{5}{2 x}\right)^{8}$ Solution: To find: $7^{\text {th }}$ term in the expansion of $\left(\frac{4 x}{5}+\frac{5}{2 x}\right)^{8}$ Formula used: (i) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$ (ii) $T_{r+1}={ }^{n} C_{r} a^{n-r} b^{r}$ For $7^{\text {th }}$ term, $\mathrm{r}+1=7$ $\Rightarrow r=6$ $\ln \left(\frac{4 x}{5}+\frac{5}{2 x}\right)^{8}$ $7^{\text {th }}$ term $=\mathrm{T}_{6+1}$ $\Rightarrow{ }^{8} \ma...

Prove that

Question: Prove that$(2+\sqrt{x})^{4}+(2-\sqrt{x})^{4}=2\left(16+24 x+x^{2}\right)$ Solution: To prove: $(2+\sqrt{x})^{4}+(2-\sqrt{x})^{4}=2\left(16+24 x+x^{2}\right)$ Formula used: (i) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$ (ii) $(a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots \ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C_{n} b^{n}$ $(a+b)^{4}={ }^{4} C_{0} a^{4}+{ }^{4} C_{1} a^{4-1} b+{ }^{4} C_{2} a^{4-2} b^{2}+{ }^{4} C_{3} a^{4-3} b^{3}+{ }^{4} C_{4}...

Using binomial theorem, prove that

Question: Using binomial theorem, prove that $\left(2^{3 n}-7 n-1\right)$ is divisible by 49, where $n$ N. Solution: To prove: $\left(2^{3 n}-7 n-1\right)$ is divisible by 49, where $n N$ Formula used: $(a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots \ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C_{n} b^{n}$ $\left(2^{3 n}-7 n-1\right)=\left(2^{3}\right)^{n}-7 n-1$ $\Rightarrow 8^{n}-7 n-1$ $\Rightarrow(1+7)^{n}-7 n-1$ $\Rightarrow{ }^{n} C_{0} 1^{n}+{ }^{n} C...

Using binominal theorem, evaluate each of the following :

Question: Using binominal theorem, evaluate each of the following : (i) $(101)^{4}$ (ii) $(98)^{4}$ (iii) $(1.2)^{4}$ Solution: (i) $(101)^{4}$ To find: Value of $(101)^{4}$ Formula used: (i) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$ (ii) $(a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots \ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C_{n} b^{n}$ $101=(100+1)$ Now $(101)^{4}=(100+1)^{4}$ $(100+1)^{4}=$ $\left[{ }^{4} C_{0}(100)^{4-0}\right]+\left[{ }^{4} C_{1}(10...

Prove that

Question: Prove that $\sum_{r=0}^{n}{ }^{n} C_{r} \cdot 3^{r}=4^{n}$ Solution: To prove $\sum_{r=0}^{n}{ }^{n} C_{r} \cdot 3^{r}=4^{n}$ Formula used: $\sum_{r=0}^{n}{ }^{n} C_{r} \cdot a^{n-r} b^{r}=(a+b)^{n}$ Proof: In the above formula if we put a = 1 and b = 3, then we will ge $\sum_{r=0}^{n}{ }^{n} C_{r} \cdot 1^{n-r} 3^{r}=(1+3)^{n}$ Therefore, $\sum_{r=0}^{n}{ }^{n} C_{r} \cdot 3^{r}=(4)^{n}$ Hence Proved....

Evaluate :

Question: Evaluate : $(\sqrt{3}+\sqrt{2})^{6}-(\sqrt{3}-\sqrt{2})^{6}$ Solution: To find: Value of $(\sqrt{3}+\sqrt{2})^{6}-(\sqrt{3}-\sqrt{2})^{6}$ Formula used: (i) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$ (ii) $(a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots \ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C_{n} b^{n}$ $(a+b)^{6}={ }^{6} C_{0} a^{6}+{ }^{6} C_{1} a^{6-1} b+{ }^{6} C_{2} a^{6-2} b^{2}+{ }^{6} C_{3} a^{6-3} b^{3}+{ }^{6} C_{4} a^{6-4} b^{4}+{ }^...

Evaluate :

Question: Evaluate : $(2+\sqrt{3})^{7}+(2-\sqrt{3})^{7}$ Solution: To find: Value of $(2+\sqrt{3})^{7}+(2-\sqrt{3})^{7}$ Formula used: (i) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$ (ii) $(a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots \ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C_{n} b^{n}$ $\begin{aligned} \left[{ }^{7} C_{0} a^{7}\right]+\left[{ }^{7} C_{1} a^{7-1} b\right]+\left[{ }^{7} C_{2} a^{7-2} b^{2}\right]+\left[{ }^{7} C_{3} a^{7-3} b^{3}\right]+\l...

Evaluate :

Question: Evaluate : $(\sqrt{3}+1)^{5}-(\sqrt{3}-1)^{5}$ Solution: To find: Value of $(\sqrt{3}+1)^{5}-(\sqrt{3}-1)^{5}$ Formula used: (I) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$ (ii) $(a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots \ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C{n} b^{n}$ $(a+1)^{5}={ }^{5} C_{0} a^{5}+{ }^{5} C_{1} a^{5-1} 1+{ }^{5} C_{2} a^{5-2} 1^{2}+{ }^{5} C_{3} a^{5-3} 1^{3}+{ }^{5} C_{4} a^{5-4} 1^{4}+{ }^{5} C_{5} 1^{5}$ $\Rightarrow...