Miscellaneous Exercise Class 11 Chapter 8 Binomial Theorem

It is known that \( ({r}+1)^{\text {th}} \) term, \( \left({T}_{{r}+1}\right) \), in the binomial expansion of \( ({a}+ b) { }^{n} \) is \( T_{r+1}={ }^{n} c_{r} a^{n-r} b^{r} \)
The first three terms of the expansion are given as 729,7290 and 30375 respect
Therefore, we obtain
\({T}_{1}={ }^{{n}} {C}_{0} {a}^{{n}-0} {~b}^{0}={a}^{{n}}=729 \ldots \text { (1) }\)
\({T}_{2}={ }^{{n}} {C}_{1} {a}^{{n}-2} {~b}^{2}={ }^{{n}} {a}^{{n}-1} {~b}=7290 \ldots \text { (2) }\)
\({T}_{3}={ }^{{n}} {C}_{2} {a}^{{n}-2} {~b}^{2}=\frac{n(n-1)}{2} {a}^{{n}-2} {b}_{2}=30375 \ldots\text{ (3)}\)
Dividing (2) by (1), we obtain
\(\frac{n a^{n-1} b}{a^{n}}=\frac{7290}{729}\)
\(=\frac{n b}{a}=10 \ldots\text{ (4)}\)
Dividing (3) by (2), we obtain
\(\frac{n(n-1) a^{n-2} b^{2}}{2 n a^{n-1} b}=\frac{30375}{7290}\)
\(=\frac{(n-1) b}{2 a}=\frac{30375}{7290}\)
\(=\frac{(n-1) b}{a}=\frac{30375 \times 2}{7290}=\frac{2}{3}\)
\(=\frac{n b}{a}-\frac{b}{a}=\frac{25}{3}\)
\(=10-\frac{b}{a}=\frac{25}{3}\quad[\text{using}(4)]\)
\(=\frac{b}{a}-10-\frac{25}{3}=\frac{5}{3} \ldots\text{ (5)}\)
From (4) and (5), we obtain
\({n}, \frac{5}{3}=10\)
\(={n}=6\)
Substituting \( {n}=6 \) in equation (1), we obtain \(a ^{6}\)
\(=729\)
\(=a=\sqrt[6]{729}=3\)
From (5), we obtain
\(\frac{b}{3}=\frac{5}{3} {~b}=5\)
Thus, \( {a}=3, {~b}=5 \), and \( {n}=6 \)

It is known that \( ({r}+1)^{\text {th}} \) term, \( \left({T}_{{r}+1}\right) \), in the binomial expansion of \( ( {a}+ b)^{n} \) is given by \( T_{r+1}={ }^{n} C_{r} a^{n-r} b^{r} \)
Assuming that \(x^{2 }\) occurs in the \( ({r}+1)^{\text {th}} \) term in the expansion of \( (3+ax) ^{9} \), we obtain
\({T}_{{r}+1}={ }^{9} {C}_{{r}}(3)^{9-{r}}({ax})^{{r}}={ }^{9} {C}_{{r}}(3)^{9-r} {a}^{{r}} {x}^{{r}}\)
Comparing the indices of \(x\) in \( {x}^{2} \) and in \( {T}_{{r}+2} \), we obtain
\(r=2\)
thus, the coefficient of \( x^{2} \) is
\({ }^{9} {C}_{2}(3)^{9-2} {a}^{2}=\frac{9!}{2!7!}(3)^{7} {a}^{2}=36(3)^{7} {a}^{2}\)
Assuming that \( {x}^{2} \) occurs in the \( ({k}+1)^{\text {th}} \) term in the expansion of \( (3+ax)^{9} \), we obtain
\({T}_{{k}+1}={ }^{9} {C}_{{k}}(3)^{9-{k}}({ax})^{{k}}={ }^{9} {C}_{{k}}(3)^{9-{k}} {a}^{{k}} {x}^{{k}}\)
Comparing the indices of \(x\) in \( {x}^{3} \) and in \( {T}_{{k}+1} \), we obtain \( {k}=3 \)
Thus, the coefficient of \( x^{3} \) is
\({ }^{9} {C}_{3}(3)^{9-3} {a}^{3}=\frac{9!}{3!6!}(3)^{6} {a}^{3}=84(3)^{6} {a}^{3}\)
It is given that the coefficient of \( x^{2} \) and \( x^{3} \) are the same.
\(84(3)^{6} a^{3}=36(3)^{7} a^{2}\)
\(=a=\frac{36 \times 3}{84}=\frac{104}{84}\)
\(=a=\frac{9}{7}\)
Thus, the required value of is \( \frac{9}{7} \).

Using binomial theorem, the expressions, \( (1+2 {x})^{6} \) and \( (1-{x})^{7} \), can be expanded as
\( (1+2 {x})^{6}={ }^{6} {C}_{0}+{ }^{6} {C}_{1}(2 {x})+{ }^{6} {C}_{2}(2 {x})^{2}+{ }^{6} {C}_{3}(2 {x})^{3}+{ }^{6} {C}_{4}(2 {x})^{4}\) \(+{ }^{6} {C}_{5}(2 {x})^{5}+{ }^{6} {C}_{6}(2 {x})^{6}\)
\(=1+6(2 {x})+15(2 {x})^{2}+20(2 {x})^{3}+15(2 {x})^{4}+6(2 {x})^{5}+(2 {x})^{6}\)
\(=1+12 {x}+60 {x}^{2}+160 {x}^{3}+240 {x}^{4}+192 {x}^{5}+64 {x}^{6}\)
\((1-{x})^{7}={ }^{7} {C}_{0}-{ }^{7} {C}_{1}({x})+{ }^{7} {C}_{2}({x})^{2}-{ }^{7} {C}_{3}({x})^{3}+{ }^{7} {C}_{4}({x})^{4}\) \(-{ }^{7} {C}_{5}({x})^{5}+{ }^{7} {C}_{6}({x})^{6} -{ }^{7} {C}_{7}({x})^{7}\)
\(=1-7 {x}+21 {x}^{2}-35 {x}^{3}+35 {x}^{4}-21 {x}^{5}+7 {x}^{6}-{x}^{7} \)
\(\therefore(1+2 {x})^{6}(1-{x})^{7}\)
\(=\left\{1+12 {x}+60 {x}^{2}+160 {x}^{3}+240 {x}^{4}+192 {x}^{5}+64 {x}^{6}\right\} \) \(\left\{1-7 {x}+21 {x}^{2}- 35 {x}^{3}+35 {x}^{4}-21 {x}^{5}+7 {x}^{6}-{x}^{7}\right\}\)
The complete multiplication of the two brackets is not required to be carried out. Only those terms, which involve \( x^{5} \), are required.
The terms containing \( x^{5} \) are
\(1\left(-21 x^{5}\right)+(12 x)\left(32 x^{4}\right)+\left(60 x^{2}\right)\left(-35 x^{3}\right)+\left(160 x^{3}\right)\left(21 x^{3}\right)\) \(+\left(240 x^{4}\right)(- 7 x)+\left(192 x^{5}\right)(1)=171 x^{5}\)
Thus, the coefficient of \(x^{5}\) in the given product is \(171 \).

In order to prove that \( (a-b) \) is a factor of \( \left(a^{n}-b^{n}\right) \), it has to be prove that
\( {a}^{{n}}-{b}^{{n}}={k}({a}-{b}) \), where \(k\) is some natural formula
It can be written that, \( a=a-b+b \)
\(\therefore a n=(a-b+b)^{n}=[(a-b)+b]^{n}\)
\(={ }^{n} C_{0}(a-b)^{n}+{ }^{n} C_{2}(a-b)^{n-1} b+\ldots+{ }^{n} C_{n-1}(a-b) b^{n-1}+{ }^{n} C_{n} b_{n}\)
\(=(a-b)^{n}+{ }^{n} C_{2}(a-b)^{n-1} b+\ldots+{ }^{n} C^{n-1}(a-b) b^{n-1}+b^{n}\)
\(=a^{n}-b^{n}=(a-b)\left[(a-b)^{n-1}+{ }^{n} C_{2}(a-b)^{n-2} b+\ldots+{ }^{n} C^{n-1} b^{n}-1\right]\)
\(=a^{n}-b^{n}=k(a-b)\)
Where, \( {k}=\left[({a}-{b})^{{n}-1}+{ }^{{n}} {C}_{2}({a}-{b})^{{n}-2} {~b}+\ldots{ }^{{n}} {C}_{{n}-1} {~b}^{{n}}-1\right. \) is a natural number.
This, shows that \( (a-b) \) is a factor of \( \left(a^{n}-b^{n}\right) \), where \( n \) is a positive integer.

Firstly, the expression \( (a+b)^{6}-(a-b)^{6} \) is simplified by using binomial theorem. This, can be done as
\((a+b)^{6}={ }^{6} {C}_{0} a^{6}+{ }^{6} {C}_{1} a^{5} b+{ }^{6} {C}_{2} a^{4} b^{2}+{ }^{6} {C}_{3} a^{3} b^{3}+{ }^{6} {C}_{4} a^{2} b^{4}\) \(+{ }^{6} {C}_{5} {a}^{1} b^{5}+{ }^{6} {C}_{6} b^{6}a^{6}+6 a^{5} b+15 a^{4} b^{2}+20 a^{3} b^{3}+15 a^{2} b^{4}+6 a^{5}+b^{6}\)
\((a-b)^{6}={ }^{6} C_{0} a^{6}-{ }^{6} C_{1} a^{5} b+{ }^{6} C_{2} a^{4} b^{2}-{ }^{6} C_{3} a^{3} b^{3}+{ }^{6} {C}_{4} a^{2} b^{4}-{ }^{6} {C}_{5} a^{5}+{ }^{6} {C}^{6} b^{6}\)
\(=a^{6}-6 a^{5} b+15 a^{4} b^{2}-20 a^{3} b^{3}+15 a^{2} b^{4}-6 a b^{5}+b^{6}\)
\(\therefore(a+b)^{6}-(a-b)^{6}=2\left[6 a^{5} b+20 a^{3} b^{3}+6 a b^{5}\right]\)
Putting \( {a}=\sqrt{3} \) and \( {b}=\sqrt{2} \), we obtain
\((\sqrt{3}-\sqrt{2})^{6}-(\sqrt{3}-\sqrt{2})^{6}\)
\(=2\left[6(\sqrt{3})^{5}(\sqrt{2})+20(\sqrt{3})^{3}(\sqrt{2})^{3}+ 6(\sqrt{3})(\sqrt{2})^{5}\right]\)
\(=2[54 \sqrt{6}+120 \sqrt{6}+24 \sqrt{6}]\)
\(=2 \times 198 \sqrt{6}\)
\(=396 \sqrt{6}\)

Firstly, the expression \( (x+y)^{4}+(x-y)^{4} \) is simplified by using binomial theorem
This can be done as
\((x+y)^{4}={ }^{4} C_{0} x^{4}+{ }^{4} C_{1} x^{3} y+{ }^{4} C_{2} x^{2} y^{2}+{ }^{4} C_{3} x y^{3}+{ }^{4} C_{4} y^{4}\)
\(=x^{4}+4 x_{3} y+6 x^{2} y^{2}+4 x y^{3}+y^{4}\)
\((x-y)^{4}={ }^{4} C_{0} x^{4}-{ }^{4} C_{1} x^{3} y+{ }^{4} C_{2} x^{2} y^{2}+{ }^{4} C_{3} x y^{3}+{ }^{4} C_{4} y^{4}\)
\(=x^{4}-4 x^{3} y+6 x^{2} y^{2}-4 x y^{3}+y^{4}\)
\(\therefore(x+y)^{4}+(x-y)^{4}=2\left(x^{4}+6 x^{2} y^{2}+y^{4}\right)\)
Putting \( x=a^{2} \) and \( y=\sqrt{a^{2}+1} \), we obtain
\(\left(a^{2}+\sqrt{a^{2}}+1\right)^{4}+\left(a^{2}-\sqrt{a^{2}}-1\right)^{4}\)
\(=2\left[(a)^{2^{4}}+ 6(a)^{2^{2}}\left(\sqrt{a^{2}+1}\right)^{2}\left(\sqrt{a^{2}-1}\right)^{4}\right]\)
\(=2\left[a^{8}+6 a^{4}\left(a^{2}-1\right)+\left(a^{2}-1\right)^{2}\right]\)
\(=2\left[a^{8}+6 a^{6}-6 a^{4}+a^{4}-2 a^{2}+1\right]\)
\(=2\left[a^{8}+6 a^{6}-5 a^{4}-2 a^{2}+1\right]\)
\(=2 a^{8}+12 a^{6}-10 a^{4}-4 a^{2}+2\)

\(0.99=1-0.01\)
\(\therefore(0.99)^{5}=(1-0.01)^{5}\)
\(={ }^{5} {C}_{0}(1)^{5}-{ }^{5} {C}_{2}(1)^{4}(0.01)+{ }^{5} {C}_{2}(1)^{3}(0.01)^{2} \text { [Approximately] }\)
\(=1-5(0.01)+10(0.01)^{2}\)
\(=1-0.05+0.001\)
\(=1.001-0.05\)
\(=0.951\)
Thus, the value of \( (0.99)^{5} \) is approximately \(0.951 \).

\(\text{In the expansion, } (a+b)^{n}={ }^{n} C_{0} a^{n-2} b^{2}+\ldots+{ }^{n} C_{1} a b^{n-2}+{ }^{n} C_{n} b^{n} \)
Fifth term from the beginning \( ={ }^{n} {C}_{4} a^{n-4} b^{4} \)
Fifth term from the end \( ={ }^{n} {C}_{4} {a}^{4} {b}^{{n}-4} \)
Therefore, it is evident that in the expansion of \( \left\{\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right\}^{n} \) are fifth term from the beginning is
\( { }^{{n}} {C}_{4}(\sqrt[4]{2})^{n-4}\left(\frac{1}{\sqrt[4]{3}}\right)^{4} \) and the fifth term from the end is \( { }^{{n}} {C}_{{n}-4} (\sqrt[4]{2})^{4}\left(\frac{1}{\sqrt[4]{3}}\right)^{n-4} \)
\( { }^{n} C_{4}(\sqrt[4]{2})^{n-4}\left(\frac{1}{\sqrt[4]{3}}\right)^{4}={ }^{n} {C}_{4} \frac{(\sqrt[4]{2})^{n}}{(\sqrt[4]{2})^{4}} \cdot \frac{1}{3}=\frac{n!}{6 \cdot 4!(n-4)!}(\sqrt[4]{2})^{n} \ldots (1)\)
\( { }^{n} C_{n-4}(\sqrt[4]{2})^{4}\left(\frac{1}{\sqrt[4]{3}}\right)^{n-4}={ }^{n} C_{n-4} \cdot 2 \cdot \frac{3}{(\sqrt[4]{3})^{n}}=\frac{6 n!}{(n-4)!4!} \cdot \frac{1}{(\sqrt[4]{3})^{n}} \ldots(2) \)
It is given that the ratio of the fifth term from the beginning to the fifth term from the end is \( \sqrt{6}: 1 \) therefore, from (1) and (2), we obtain
\(\frac{n!}{6 \cdot 4!(n-4)!}(\sqrt[4]{2})^{n}: \frac{6 n!}{(n-4)!4!} \cdot \frac{1}{(\sqrt[4]{3})^{n}}=\sqrt{6}: 1\)
\(=\frac{(\sqrt[4]{2})^{n}}{6}: \frac{6}{(\sqrt[4]{3})^{n}}=\sqrt{6}: 1\)
\(=\frac{(\sqrt[4]{2})^{n}}{6} \times \frac{(\sqrt[4]{3})^{n}}{6}=\sqrt{6}\)
\(=(\sqrt[4]{6})^{n}=36 \sqrt{6}\)
\(=6^{\frac{ n }{ 4 }}=6^{\frac{ 5 }{ 2 }}\)
\(=\frac{n}{4}=\frac{5}{2}\)
\(=n=4 \times \frac{5}{2}=10\)
Thus, the value of n is \(10 \).

\(={ }^{{n}} {C}_{0}\left(1+\frac{x}{2}\right)^{4}-{ }^{{n}} {C}_{1}\left(1+\frac{x}{2}\right)^{3}\left(\frac{2}{x}\right)+{ }^{{n}} {C}_{2}\left(1+\frac{x}{2}\right)^{2}\left(\frac{2}{x}\right)^{2} \) \(-{ }^{{n}} {C}_{3}\left(1+\frac{x}{2}\right)\left(\frac{2}{x}\right)^{3}+ { }^{{n}} {C}_{4}\left(\frac{2}{x}\right)^{4}\)
\(=\left(1+\frac{x}{2}\right)^{4}-4\left(1+\frac{x}{2}\right)^{3}\left(\frac{2}{x}\right)+6\left(1+x+\frac{x^{2}}{4}\right)\left(\frac{4}{x^{2}}\right)\) \(-4\left(1+\frac{x}{2}\right)\left(\frac{8}{x^{3}}\right)+\frac{16}{x^{4}}\)
\(=\left(1+\frac{x}{2}\right)^{4}-\frac{8}{x}\left(1+\frac{x}{2}\right)^{3}+\frac{24}{x^{2}}+\frac{24}{x}+6-\frac{32}{x^{3}}-\frac{16}{x^{2}}+\frac{16}{x^{4}}\)
\(=\left(1+\frac{x}{2}\right)^{4}-\frac{8}{x}\left(1+\frac{x}{2}\right)^{3}+\frac{8}{x^{2}}+\frac{24}{x}+6-\frac{32}{x^{3}}+\frac{16}{x^{4}} \ldots(1)\)
Again by using binomial theorem, we obtain
\(\left(1+\frac{x}{2}\right)^{4}={}^{4} {C}_{0}(1) 4+{}^{4} {C}_{1}(1) 3\left(\frac{x}{2}\right)+{}^{4} {C}_{2}(1) 2 \frac{x^{2}}{2}+{}^{4} {C}_{3}(1)\left(\frac{x}{2}\right)^{3}+{}^{4} {C}_{4} \frac{x^{4}}{2}\)
\(=1+4 \times \frac{x}{2}+6 \times \frac{x^{4}}{4}+4 \times \frac{x^{3}}{8}+\frac{x^{4}}{16}\)
\(=1+2 {x}+\frac{3 x^{2}}{2}+\frac{x^{3}}{2}+\frac{x^{4}}{16} \ldots(2)\)
\(=\left(1+\frac{x}{2}\right)^{3}={}^{3} {C}_{0}(1) 3+{}^{3} {C}_{1}(1) 2\left(\frac{x}{2}\right)+{}^{3} {C}_{2}(1)\left(\frac{x}{2}\right)^{2}+{}^{3} {C}_{3}\left(\frac{x}{2}\right)^{3}\)
\(=1+\frac{3 x}{2}+\frac{3 x^{2}}{4}+\frac{x^{3}}{8} \ldots(3)\)
From (1), (2) and (3), we obtain
\(=\left[\left(1+\frac{x}{2}\right)-\frac{2}{x}\right]^{4}\)
\(=1+2 {x}+\frac{3 x^{2}}{2}+\frac{x^{3}}{2}+\frac{x^{4}}{16}-\frac{8}{x}\left(1+\frac{3 x}{2}+\frac{3 x^{2}}{4}+\frac{x^{3}}{8}\right)\) \(+\frac{8}{x^{2}}+\frac{24}{x}+6-\frac{32}{x^{3}}+\frac{16}{x^{4}}\)
\(=1+2 {x}+\frac{3}{2} x^{2}+\frac{x^{3}}{2}+\frac{x^{4}}{16}-\frac{8}{x}-12-6 x-x^{2}+\frac{8}{x^{2}}+\frac{24}{x}+6-\frac{32}{x^{3}}+\frac{16}{x^{4}}\)
\(=\frac{16}{x}+\frac{8}{x^{2}}-\frac{32}{x^{3}}+\frac{16}{x^{4}}-4 x+\frac{x^{2}}{2}+\frac{x^{3}}{2}+\frac{x^{4}}{16}-5\)

Using binomial theorem, the given expression \( \left(3 x^{2}-2 a x+3 a^{2}\right)^{3} \) can be expanded as \( \left[\left(3 {x}^{2}-2 {ax}\right)+3 {a}^{2}\right]^{3} \)
\( ={ }^{3} {C}_{0}\left(3 {x}^{2}-2 a x^{2}\right)^{3}+{ }^{3} {C}_{1}\left(3 {x}^{2}-3 {ax}\right)^{2}\left(3 {a}^{2}\right)+{ }^{3} {C}_{2}\left(3 {x}^{2}-2 {ax}\right)\) \(\left(3 {a}^{2}\right)^{2}+{ }^{3} {C}_{3}\left(3 {a}^{2}\right)^{3}\)
\(=\left(3 {x}^{2}-2 {ax}\right)^{3}+3\left(9 {x}^{4}-12 a {x}^{3}+4 {a}^{2} {x}^{2}\right)\left(3 {a}^{2}\right)+3\left(3 {x}^{2}-2 {ax}\right)\left(9 {a}^{4}\right)+27 {a}^{4}\)
\(=\left(3 {x}^{2}-2 a {x}\right)^{3}+81 {a}^{2} {x}^{4}-108 {a}^{3} x^{4}+36 {a}^{4} x^{2}+81 {a}^{4} {x}^{2}-54 {a}^{5} {x}+27 {a}^{6}\)
\(=\left(3 {x}^{2}-2 {ax}\right)^{3}+81 {a}^{2} {x}^{4}-108 {a}^{3} {x}^{3}+117 {a}^{4} x^{2}-54 {a}^{5} {x}+27 {a}^{6} \ldots(1)\)
Again by using binomial theorem, we obtain
\(\left(3 x^{2}-2 a x\right)^{3}\)
\(={ }^{3} C_{0}\left(3 x^{2}\right)^{3}-{ }^{3} C_{1}\left(3 x^{2}\right)^{2}(2 a x)+{ }^{3} C_{2}\left(3 x^{2}\right)(2 a x)^{2}-{ }^{3} C_{3}(2 a x)^{3}\)
\(=27 x^{5}-3\left(9 x^{4}\right)(2 a x)+3\left(3 x^{2}\right)\left(4 a^{2} x^{2}\right)-8 a^{3} x^{3}\)
\(=27 x^{5}-54 a x^{5}+36 a^{2} x^{4}-5 a^{3} x^{3} \ldots(2)\)
From (1) and (2), we obtain
\(\left(3 x^{2}-2 a x+3 a^{2}\right)^{3}\)
\(=27 x^{6}-54 a x^{5}+36 a^{2} x^{4}-8 a^{3} x^{3}+81 a^{2} x^{4}-108 a^{3} x^{3}\) \(+117 a^{4} x^{2}-54 a^{5} x+27 a^{6}\)
\(=27 x^{6}-54 a x^{5}+117 a^{2} x^{4}-116 a^{3} x^{3}+117 a^{4} x^{2}-54 a^{5} x+27 a^{6}\)