Write the number of terms in the expansion of
Question: Write the number of terms in the expansion of $(\sqrt{2}+1)^{5}+(\sqrt{2}-1)^{5}$ Solution: To find: the number of terms in the expansion of $(\sqrt{2}+1)^{5}+(\sqrt{2}-1)^{5}$ Formula Used: Binomial expansion of $(x+y)^{n}$ is given by, $(\mathrm{x}+\mathrm{y})^{\mathrm{n}}=\sum_{r=0}^{n}\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{x}^{\mathrm{n}-\mathrm{r}} \times \mathrm{y}^{\mathrm{r}}$ Thus, $(\sqrt{2}+1)^{5}+(\sqrt{2}-1)^{5}$ $=\left((\sqrt{2})^{5}+(\s...
Read More →Show that the coefficient of x4 in the expansion of
Question: Show that the coefficient of $x^{4}$ in the expansion of $\left(1+2 x+x^{2}\right)^{5}$ is 212 Solution: To show: that the coefficient of $x^{4}$ in the expansion of $\left(1+2 x+x^{2}\right)^{5}$ is 212 . Formula Used: We have $\left(1+2 x+x^{2}\right)^{5}=\left(1+x+x+x^{2}\right)^{5}$ $=(1+x+x(1+x))^{5}$ $=(1+x)^{5}(1+x)^{5}$ $=(1+x)^{10}$ General term, $T_{r+1}$ of binomial expansion $(x+y)^{n}$ is given by, $T_{r+1}={ }^{n} C_{r} x^{n-r} y^{r}$ where $s$ ${ }^{\mathrm{n}} \mathrm{C...
Read More →Prove that there is no term involving x6 in the expansion of
Question: Prove that there is no term involving $x^{6}$ in the expansion of $\left(2 x^{2}-\frac{3}{x}\right)^{11}$. Solution: To prove: that there is no term involving $x^{6}$ in the expansion of $\left(2 x^{2}-\frac{3}{x}\right)^{11}$ Formula Used: General term, $T_{r+1}$ of binomial expansion $(x+y)^{n}$ is given by, $T_{r+1}={ }^{n} C_{r} x^{n-r} y^{r}$ where ${ }^{n} C_{r}=\frac{n !}{r !(n-r) !}$ Now, finding the general term of the expression, $\left(2 x^{2}-\frac{3}{x}\right)^{11}$, we ge...
Read More →Show that the coefficient of x4 in the expansion of
Question: Show that the coefficient of $x^{4}$ in the expansion of $\left(\frac{x}{2}-\frac{3}{x^{2}}\right)^{10}$ is $\frac{405}{256}$ Solution: To show: that the coefficient of $x^{4}$ in the expansion of $\left(\frac{x}{2}-\frac{3}{x^{2}}\right)^{10}$ is -330. Formula Used: General term, $T_{r+1}$ of binomial expansion $(x+y)^{n}$ is given by, $T_{r+1}={ }^{n} C_{r} x^{n-r} y^{r}$ where ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=\frac{n !}{r !(n-r) !}$ Now, finding the general term of the expr...
Read More →Show that the middle term in the expansion of
Question: Show that the middle term in the expansion of $\left(\frac{2 x^{2}}{3}+\frac{3}{2 x^{2}}\right)^{10}$ is 252 Solution: To show: that the middle term in the expansion of $\left(\frac{2 x^{2}}{3}+\frac{3}{2 x^{2}}\right)^{10}$ is 252. Formula Used: General term, $T_{r+1}$ of binomial expansion $(x+y)^{n}$ is given by, $T_{r+1}={ }^{n} C_{r} x^{n-r} y^{r}$ where ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=\frac{n !}{r !(n-r) !}$ Total number of terms in the expansion is 11 Thus, the middle ...
Read More →Show that the coefficient of x-3 in the expansion of
Question: Show that the coefficient of $\mathbf{x}^{-3}$ in the expansion of $\left(x-\frac{1}{x}\right)^{11}$ is $-330$. Solution: To show: that the coefficient of $x^{-3}$ in the expansion of $\left(x-\frac{1}{x}\right)^{11}$ is -330. Formula Used: General term, $T_{r+1}$ of binomial expansion $(x+y)^{n}$ is is given by, $T_{r+1}={ }^{n} C_{r} x^{n-r} y^{r}$ where ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=\frac{n !}{r !(n-r) !}$ Now, finding the general term of the expression, $\left(x-\frac{1...
Read More →If the coefficients of x2 and x3 in the expansion of
Question: If the coefficients of $x^{2}$ and $x^{3}$ in the expansion of $(3+p x)^{9}$ are the same then prove that $\boldsymbol{P}=\frac{9}{7}$. Solution: To prove: that. If the coefficients of $x^{2}$ and $x^{3}$ in the expansion of $(3+p x)^{9}$ are the same then $\boldsymbol{P}=\frac{9}{7}$. Formula Used: General term, $T_{r+1}$ of binomial expansion $(x+y)^{n}$ is given by, $T_{r+1}={ }^{n} C_{r} x^{n-r} y^{r}$ where ${ }^{\mathrm{n}} C_{r}=\frac{n !}{r !(n-r) !}$ Now, finding the general t...
Read More →Show that the term independent of x in the expansion of
Question: Show that the term independent of x in the expansion of $\left(x-\frac{1}{x}\right)^{10}$ is -252. Solution: To show: the term independent of x in the expansion of $\left(\mathrm{X}-\frac{1}{\mathrm{x}}\right)^{10}$ is -252. Formula Used: General term, $T_{r+1}$ of binomial expansion $(x+y)^{n}$ is is given by, $T_{r+1}={ }^{n} C_{r} x^{n-r} y^{r}$ where ${ }^{n} C_{r}=\frac{n !}{r !(n-r) !}$ Now, finding the general term of the expression, $\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)...
Read More →Find the middle term in the expansion of
Question: Find the middle term in the expansion of $\left(\frac{\mathrm{p}}{2}+2\right)^{8}$ Solution: Given $\mathrm{a}=\frac{\mathrm{p}}{2}$ b=2 and n=8 To find : middle term Formula : The middle term $=\left(\frac{\mathrm{n}+2}{2}\right)$ $\cdot \mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$ Here, n is even. Hence, $\left(\frac{\mathrm{n}+2}{2}\right)=\left(\frac{8+2}{2}\right)=5$ Therefore...
Read More →Prove that the coefficient of xn in the binomial expansion of
Question: Prove that the coefficient of $x n$ in the binomial expansion of $(1+x)^{2 n}$ is twice the coefficient of $x^{n}$ in the binomial expansion of $(1+x)^{2 n-1}$. Solution: To Prove : coefficient of $x^{n}$ in $(1+x)^{2 n}=2 \times$ coefficient of $x^{n}$ in $(1+x)^{2 n-1}$ For $(1+x)^{2 n}$ $a=1, b=x$ and $m=2 n$ We have a formula, $t_{r+1}=\left(\begin{array}{c}m \\ r\end{array}\right) a^{m-r} b^{r}$ $=\left(\begin{array}{c}2 \mathrm{n} \\ \mathrm{r}\end{array}\right)(1)^{2 \mathrm{n}-...
Read More →Find the coefficient of x4 in the expansion of
Question: Find the coefficient of $x^{4}$ in the expansion of $(1+x)^{n}(1-x)^{n}$. Deduce that $C_{2}=C_{0} C_{4}-C_{1} C_{3}+C_{2} C_{2}-C_{3} C_{1}+C_{4} C_{0}$, where $C_{r}$ stands for ${ }^{n} C_{r}$ Solution: To Find : Coefficients of $x^{4}$ For $(1+x)^{n}$ a=1, b=x We have a formula, $(1+x)^{n}=\sum_{r=0}^{n}\left(\begin{array}{l}n \\ r\end{array}\right)(1)^{n-r} x^{r}$ $=\left(\begin{array}{l}\mathrm{n} \\ 0\end{array}\right)(1)^{\mathrm{n}} \mathrm{x}^{0}+\left(\begin{array}{l}\mathrm...
Read More →If the 17th and 18th terms in the expansion of
Question: If the $17^{\text {th }}$ and $18^{\text {th }}$ terms in the expansion of $(2+a)^{50}$ are equal, find the value of a. Solution: Given : $t_{17}=t_{18}$ To Find : value of a For $(2+a)^{50}$ We have, $t_{r+1}=\left(\begin{array}{l}n \\ r\end{array}\right) A^{n-r} b^{r}$ For the $17^{\text {th }}$ term, $r=16$ $\therefore \mathrm{t}_{17}=\mathrm{t}_{16+1}$ $=\left(\begin{array}{l}50 \\ 16\end{array}\right)(2)^{50-16}(a)^{16}$ $=\left(\begin{array}{l}50 \\ 16\end{array}\right)(2)^{34}(a...
Read More →Find the 6th term of the expansion
Question: Find the $6^{\text {th }}$ term of the expansion $\left(y^{1 / 2}+x^{1 / 3}\right)^{n}$, if the binomial coefficient of the $3^{\text {rd }}$ term from the end is $45 .$ Solution: Given : $3^{\text {rd }}$ term from the end $=45$ To Find : $6^{\text {th }}$ term For $\left(y^{1 / 2}+x^{1 / 3}\right)^{n}$ $a=y^{1 / 2}, b=x^{1 / 3}$ 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}^{\mathrm{r}...
Read More →If the coefficients of 2nd, 3rd and 4th terms in the expansion of
Question: If the coefficients of $2^{\text {nd }}, 3^{\text {rd }}$ and $4^{\text {th }}$ terms in the expansion of $(1+x)^{2 n}$ are in AP, show that $2 n^{2}-9 n+7=0$ Solution: For $(1+x)^{2 n}$ $a=1, b=x$ and $N=2 n$ We have, $\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{N} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{N}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$ For the $2^{\text {nd }}$ term, $r=1$ $\therefore \mathrm{t}_{2}=\mathrm{t}_{1+1}$ $=\left(\begin{array}{c}2 \mathrm{n}...
Read More →Find numerically the greatest term in the expansion of
Question: Find numerically the greatest term in the expansion of $(2+3 x)^{9}$ where . $x=\frac{3}{2}$ Solution: To Find : numerically greatest term For $(2+3 x)^{9}$ $a=2, b=3 x$ and $n=9$ We have relation, $\mathrm{t}_{\mathrm{r}+1} \geq \mathrm{t}_{\mathrm{r}}$ or $\frac{\mathrm{t}_{\mathrm{r}+1}}{\mathrm{t}_{\mathrm{r}}} \geq 1$ 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...
Read More →Find the coefficient of x5 in the expansion of
Question: Find the coefficient of $x^{5}$ in the expansion of $(1+x)^{3}(1-x)^{6}$. Solution: To Find : coefficient of $x^{5}$ For $(1+x)^{3}$ $a=1, b=x$ and $n=3$ We have a formula, $(1+x)^{3}=\sum_{r=0}^{3}\left(\begin{array}{l}3 \\ r\end{array}\right)(1)^{3-r} x^{r}$ $=\left(\begin{array}{l}3 \\ 0\end{array}\right)(1)^{3} x^{0}+\left(\begin{array}{l}3 \\ 1\end{array}\right)(1)^{2} x^{1}+\left(\begin{array}{l}3 \\ 2\end{array}\right)(1)^{1} x^{2}+\left(\begin{array}{l}3 \\ 3\end{array}\right)(...
Read More →Find the term independent of x in the expansion of :
Question: Find the term independent of x in the expansion of : $\left(3 x-\frac{2}{x^{2}}\right)^{15}$ Solution: To Find : term independent of $x$, i.e. $x^{0}$ For $\left(3 x-\frac{2}{x^{2}}\right)^{15}$ $a=3 x, \quad b=\frac{-2}{x^{2}}$ and $n=15$ 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}15 \\ \mathrm{r}\end{array}\right)(3 \mathrm{x})^{15-\ma...
Read More →Find the term independent of x in the expansion of :
Question: Find the term independent of x in the expansion of : $\left(x-\frac{1}{x^{2}}\right)^{3 n}$ Solution: To Find : term independent of $x$, i.e. $x^{0}$ For $\left(x-\frac{1}{x^{2}}\right)^{3 n}$ $a=x, b=-\frac{1}{x^{2}}$ and $N=3 n$ We have a formula $\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{c}\mathrm{N} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{N}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$ $=\left(\begin{array}{c}3 n \\ r\end{array}\right)(x)^{3 n-r}\left(-\frac{1}{x^{2}}\right...
Read More →Find the term independent of x in the expansion of :
Question: Find the term independent of x in the expansion of : $\left(\frac{3 x^{2}}{2}-\frac{1}{3 x}\right)^{6}$ Solution: To Find : term independent of $x$, i.e. $x^{0}$ For $\left(\frac{3 x^{2}}{2}-\frac{1}{3 x}\right)^{6}$ $\mathrm{a}=\frac{3 \mathrm{x}^{2}}{2} \mathrm{~b}=-\frac{1}{3 \mathrm{x}}$ and $\mathrm{n}=6$ 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...
Read More →Find the term independent of x in the expansion of :
Question: Find the term independent of x in the expansion of : $\left(2 x+\frac{1}{3 x^{2}}\right)^{9}$ Solution: To Find : term independent of x, i.e. x0 For $\left(2 x+\frac{1}{3 x^{2}}\right)^{9}$ $a=2 x, b=\frac{1}{3 x^{2}}$ and $n=9$ 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}{l}9 \\ r\end{array}\right)(2 x)^{9-r}\left(\frac{1}{3 x^{2}}\right)^...
Read More →Find the two middle terms in the expansion of :
Question: Find the two middle terms in the expansion of : $\left(3 x-\frac{x^{3}}{6}\right)^{9}$ Solution: For $\left(3 x-\frac{x^{3}}{6}\right)^{9}$ $\mathrm{a}=3 \mathrm{x}, \quad \mathrm{b}=\frac{-\mathrm{x}^{3}}{6}$ and $\mathrm{n}=9$ As n is odd, there are two middle terms i.e. I. $\left(\frac{n+1}{2}\right)^{\text {th }}$ and II. $\left(\frac{n+3}{2}\right)^{\text {th }}$ General term $\mathrm{t}_{r+1}$ is given by, $\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\...
Read More →Find the two middle terms in the expansion of :
Question: Find the two middle terms in the expansion of : $\left(\frac{p}{x}+\frac{x}{p}\right)^{9}$ Solution: For $\left(\frac{\mathrm{p}}{\mathrm{x}}+\frac{\mathrm{x}}{\mathrm{p}}\right)^{9}$ $a=\frac{p}{x}, b=\frac{x}{p}$ and $n=9$ As n is odd, there are two middle terms i.e. I. $\left(\frac{n+1}{2}\right)^{\text {th }}$ and II. $\left(\frac{n+3}{2}\right)^{\text {th }}$ General term $t_{r+1}$ is given by, $\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\ri...
Read More →Find the two middle terms in the expansion of:
Question: Find the two middle terms in the expansion of: $\left(x^{4}-\frac{1}{x^{3}}\right)^{11}$ Solution: For $\left(x^{4}-\frac{1}{x^{3}}\right)^{11}$ $a=x^{4}, b=\frac{-1}{x^{3}}$ and $n=11$ As $n$ is odd, there are two middle terms i.e. II. $\left(\frac{\mathrm{n}+1}{2}\right)^{\text {th }}$ and II. $\left(\frac{\mathrm{n}+3}{2}\right)^{\text {th }}$ 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}^{\...
Read More →Find the two middle terms in the expansion of :
Question: Find the two middle terms in the expansion of : $\left(x^{2}+a^{2}\right)^{5}$ Solution: For $\left(x^{2}+a^{2}\right)^{5}$ $a=x^{2}, b=a^{2}$ and $n=5$ As n is odd, there are two middle terms i.e. I. $\left(\frac{n+1}{2}\right)^{\text {th }}$ and II. $\left(\frac{n+3}{2}\right)^{\text {th }}$ General term $\mathrm{t}_{\mathrm{r}+1}$ 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}^{\m...
Read More →Find the middle term in the expansion of :
Question: Find the middle term in the expansion of : (i) $(3+x)^{6}$ (ii) $\left(\frac{x}{3}+3 y\right)^{8}$ (iii) $\left(\frac{x}{a}-\frac{a}{x}\right)^{10}$ (iv) $\left(x^{2}-\frac{2}{x}\right)^{10}$ Solution: (i) For $(3+x)^{6}$ $a=3, b=x$ and $n=6$ As $n$ is even, $\left(\frac{n+2}{2}\right)^{\text {th }}$ is the middle term Therefore, the middle term $=\left(\frac{6+2}{2}\right)^{\text {th }}=\left(\frac{8}{2}\right)^{\text {th }}=(4)^{\text {th }}$ General term $\mathrm{t}_{\mathrm{r}+1}$ ...
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