NCERT Solutions for Miscellaneous Exercise Chapter 8 Class 11 - Binomial Theorem

Binomial Theorem Class 11 Chapter 8 - Miscellaneous Exercise

Question:1 Find a, b and n in the expansion of $(a + b)^n$ if the first three terms of the expansion are 729, 7290 and 30375, respectively.

Answer:

As we know the Binomial expansion of $(a + b)^n$ is given by

$(a+b)^n=^nC_0a^n+^nC_1a^{n-1}b+^nC_2a^{n-2}b^2+......^nC_nb^n$

Given in the question,

$^nC_0a^n=729.......(1)$

$^nC_1a^{n-1}b=7290.......(2)$

$^nC_2a^{n-2}b^2=30375.......(3)$

Now, dividing (1) by (2) we get,

$\Rightarrow \frac{^nC_0a^n}{^nC_1a^{n-1}b}=\frac{729}{7290}$

$\Rightarrow \frac{\frac{n!}{n!0!}}{\frac{n!}{1!(n-1)!}}\times\frac{a}{b}=\frac{729}{7290}$

$\Rightarrow\frac{(n-1)!}{n!}\times\frac{a}{b}=\frac{1}{10}$

$\Rightarrow\frac{1}{n}\times\frac{a}{b}=\frac{1}{10}$

$10a=nb......(4)$

Now, Dividing (2) by (3) we get,

$\Rightarrow \frac{^nC_1a^{n-1}b}{^nC_2a^{n-2}b^2}=\frac{7290}{30375}$

$\Rightarrow \frac{\frac{n!}{1!(n-1)!}}{\frac{n!}{2!(n-2)!}}\times\frac{a}{b}=\frac{7290}{30375}$

$\Rightarrow \frac{2(n-2)!}{(n-1)!}\times\frac{a}{b}=\frac{7290}{30375}$

$\Rightarrow \frac{2}{(n-1)}\times\frac{a}{b}=\frac{7290}{30375}$

$\Rightarrow 2\times30375\times a=7290\times b\times(n-1)$

$\Rightarrow 60750a=7290b(n-1).......(5)$

Now, From (4) and (5), we get,

$n=6,a=3\:and\:b=5$

Question:2 Find a if the coefficients of $x^2$and $x^3$ in the expansion of $(3 + ax)^9$ are equal.

Answer:

As we know that the general $(r+1)^{th}$ term $T_{r+1}$ in the binomial expansion of $(a+b)^n$ is given by

$T_{r+1}=^nC_ra^{n-r}b^r$

So, the general $(r+1)^{th}$ term $T_{r+1}$ in the binomial expansion of $(3 + ax)^9$ is

$T_{r+1}=^nC_r3^{n-r}(ax)^r=^nC_r3^{n-r}a^rx^r$

Now, $x^2$ will come when $r=2$ and $x^3$ will come when $r=3$

So, the coefficient of $x^2$ is

$K_{x^2}=^nC_23^{9-2}a^2=^nC_23^7a^2$

And the coefficient of $x^3$ is

$K_{x^3}=^9C_33^{9-3}a^2=^9C_33^6a^3$

Now, Given in the question,

$K_{x^2}=K_{x^3}$

$^9C_23^7a^2=^9C_33^6a^3$

$\frac{9!}{2!7!}\times3=\frac{9!}{3!6!}\times a$

$a=\frac{18}{14}=\frac{9}{7}$

Hence the value of a is 9/7.

Question:3 Find the coefficient of $x^5$ in the product $(1 + 2x)^6 (1 - x)^7$ using binomial theorem.

Answer:

First, lets expand both expressions individually,

So,

$(1+2x)^6=^6C_0+^6C_1(2x)+^6C_2(2x)^2+^6C_3(2x)^3+^6C_4(2x)^4+^6C_5(2x)^5+$$^6C_6(2x)^6$

$(1+2x)^6=^6C_0+2\times^6C_1x+4\times^6C_2x^2+8\times^6C_3x^3+16\times^6C_4x^4+32\times^6C_5x^5+$$64\times^6C_6x^6$

$(1+2x)^6=1+12x+60x^2+160x^3+240x^4+192x^5+64x^6$

And

$(1-x)^7=^7C_0-^7C_1x+^7C_2x^2-^7C_3x^3+^7C_4x^4-^7C_5x^5+^7C_6x^6-^7C_7x^7$

$(1-x)^7=1-7x+21x^2-35x^3+35x^4-21x^5+7x^6-x^7$

Now,

$(1 + 2x)^6 (1 - x)^7=(1+12x+60x^2+160x^3+240x^4+192x^5+64x^6)$$(1-7x+21x^2-35x^3+35x^4-21x^5+7x^6-x^7)$

Now, for the coefficient of $x^5$, we multiply and add those terms whose product gives $x^5$.So,

The term which contain $x^5$are,

$\Rightarrow (1)(-21x^5)+(12x)(35x^4)+(60x^2)(-35x^3)+(160x^3)(21x^2)+(240x^4)(-7x)$$+(192x^5)(1)$

$\Rightarrow 171x^5$

Hence the coefficient of $x^5$ is 171.

Question:4 If a and b are distinct integers, prove that$a - b$ is a factor of $a^n - b^n$ , whenever n is a positive integer.
[Hint: write $a^n = (a - b + b)^n$ and expand]

Answer:

we need to prove,

$a^n-b^n=k(a-b)$ where k is some natural number.

Now let's add and subtract b from a so that we can prove the above result,

$a=a-b+b$

$a^n=(a-b+b)^n=[(a-b)+b]^n$

$=^nC_0(a-b)^n+^nC_1(a-b)^{n-1}b+........^nC_nb^n$

$=(a-b)^n+^nC_1(a-b)^{n-1}b+........^nC_{n-1}(a-b)b^{n-1}+b^n$$\Rightarrow a^n-b^n=(a-b)[(a-b)^{n-1}+^nC_2(a-b)^{n-2}+........+^nC_{n-1}b^{n-1}]$

$\Rightarrow a^n-b^n=k(a-b)$

Hence,$a - b$ is a factor of $a^n - b^n$.

Question:5 Evaluate $\left(\sqrt3 + \sqrt2 \right )^6 - \left(\sqrt{3} - \sqrt2 \right )^6$.

Answer:

First let's simplify the expression $(a+b)^6-(a-b)^6$ using binomial theorem,

So,

$(a+b)^6=^6C_0a^6+^6C_1a^5b+^6C_2a^4b^2+^6C_3a^3b^3+^6C_4a^2b^4+^6C_5ab^5+^6C_6b^6$$(a+b)^6=a^6+6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6$

And

$(a-b)^6=^6C_0a^6-^6C_1a^5b+^6C_2a^4b^2-^6C_3a^3b^3+^6C_4a^2b^4-^6C_5ab^5+^6C_6b^6$

$(a+b)^6=a^6-6a^5b+15a^4b^2-20a^3b^3+15a^2b^4-6ab^5+b^6$

Now,

$(a+b)^6-(a-b)^6=a^6+6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6$$-a^6+6a^5b-15a^4b^2+20a^3b^3-15a^2b^4+6ab^5-b^6$

$(a+b)^6-(a-b)^6=2[6a^5b+20a^3b^3+6ab^5]$

Now, Putting $a=\sqrt{3}\:and\:b=\sqrt{2},$ we get

$\left(\sqrt3 + \sqrt2 \right )^6 - \left(\sqrt{3} - \sqrt2 \right )^6=2[6(\sqrt{3})^5(\sqrt{2})+20(\sqrt{3})^3(\sqrt{2})^3+6(\sqrt{3})(\sqrt{2})^5]$

$\left(\sqrt3 + \sqrt2 \right )^6 - \left(\sqrt{3} - \sqrt2 \right )^6=2[54\sqrt{6}+120\sqrt{6}+24\sqrt{6}]$

$\left(\sqrt3 + \sqrt2 \right )^6 - \left(\sqrt{3} - \sqrt2 \right )^6=2\times198\sqrt{6}$

$\left(\sqrt3 + \sqrt2 \right )^6 - \left(\sqrt{3} - \sqrt2 \right )^6=396\sqrt{6}$

Question:6 Find the value of $\left(a^2 + \sqrt{a^2 -1} \right )^4 + \left(a^2 - \sqrt{a^2 -1} \right )^4$

Answer:

First, lets simplify the expression $(x+y)^4-(x-y)^4$ using binomial expansion,

$(x+y)^4=^4C_0x^4+^4C_1x^3y+^4C_2x^2y^2+^4C_3xy^3+^4C_4y^4$

$(x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4$

And

$(x-y)^4=^4C_0x^4-^4C_1x^3y+^4C_2x^2y^2-^4C_3xy^3+^4C_4y^4$

$(x-y)^4=x^4-4x^3y+6x^2y^2-4xy^3+y^4$

Now,

$(x+y)^4-(x-y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4-$$x^4+4x^3y-6x^2y^2+4xy^3-y^4$

$(x+y)^4-(x-y)^4=2(x^4+6x^2y^2+y^4)$

Now, Putting $x=a^2\and\:y=\sqrt{a^2-1}$ we get,

$\left(a^2 + \sqrt{a^2 -1} \right )^4 + \left(a^2 - \sqrt{a^2 -1} \right )^4=2[(a^2)^4+6(a^2)^2(\sqrt{a^2-1})^2+(\sqrt{a^2-1})^4]$

$\left(a^2 + \sqrt{a^2 -1} \right )^4 + \left(a^2 - \sqrt{a^2 -1} \right )^4=2[a^8+6a^4(a^2-1)+(a^2-1)^2]$

$\left(a^2 + \sqrt{a^2 -1} \right )^4 + \left(a^2 - \sqrt{a^2 -1} \right )^4=2a^8+12a^6-12a^4+2a^4-4a^2+2$

$\left(a^2 + \sqrt{a^2 -1} \right )^4 + \left(a^2 - \sqrt{a^2 -1} \right )^4=2a^8+12a^6-10a^4-4a^2+2$

Question:7 Find an approximation of (0.99)⁵ using the first three terms of its expansion.

Answer:

As we can write 0.99 as 1-0.01,

$(0.99)^5=(1-0.001)^5=^5C_0(1)^5-^5C_1(1)^4(0.01)+^5C_2(1)^3(0.01)^2$$+\:other \:negligible \:terms$

$\Rightarrow (0.99)^5=1-5(0.01)+10(0.01)^2$

$\Rightarrow (0.99)^5=1-0.05+0.001$

$\Rightarrow (0.99)^5=0.951$

Hence the value of $(0.99)^5$ is 0.951 approximately.

Question:8 Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of $\left(\sqrt[4]{2} + \frac{1}{\sqrt[4]{3}} \right )^n$is $\sqrt6 :1$

Answer:

Given, the expression

$\left(\sqrt[4]{2} + \frac{1}{\sqrt[4]{3}} \right )^n$

Fifth term from the beginning is

$T_5=^nC_4(\sqrt[4]{2})^{n-4}\left(\frac{1}{\sqrt[4]{3}}\right)^4$

$T_5=^nC_4\frac{(\sqrt[4]{2})^n}{(\sqrt[4]{2})^4}\times\frac{1}{3}$

$T_5=\frac{n!}{4!(n-4)!}\times\frac{(\sqrt[4]{2})^n}{2}\times\frac{1}{3}$

And Fifth term from the end is,

$T_{n-5}=^nC_{n-4}(\sqrt[4]{2})^4\left ( \frac{1}{\sqrt[4]{3}} \right )^{n-4}$

$T_{n-5}=^nC_{n-4}(\sqrt[4]{2})^4\left ( \frac{(\sqrt[4]{3})^4}{(\sqrt[4]{3})^n} \right )$

$T_{n-5}=\frac{n!}{4!(n-4)!}\times2\times\left ( \frac{3}{(\sqrt[4]{3})^n} \right )$

Now, As given in the question,

$T_5:T_{n-5}=\sqrt{6}:1$

So,

$\left(\frac{n!}{4!(n-4)!}\times\frac{(\sqrt[4]{2})^n}{2}\times\frac{1}{3}\right):\left( \frac{n!}{4!(n-4)!}\times2\times\left ( \frac{3}{(\sqrt[4]{3})^n} \right )\right)=\sqrt{6}:1$

From Here ,

$\frac{(\sqrt[4]{2})^n}{6}:\frac{6}{(\sqrt[4]{3})^n}=\sqrt{6}:1$

$\frac{(\sqrt[4]{2})^n(\sqrt[4]{3})^n}{6\times6}=\sqrt{6}$

$(\sqrt[4]{6})^n=36\sqrt{6}$

$6^{\frac{n}{4}}=6^{\frac{5}{2}}$

From here,

$\frac{n}{4}=\frac{5}{2}$

$n=10$

Hence the value of n is 10.

Question:9 Expand using Binomial Theorem $\left(1 + \frac{x}{2} - \frac{2}{x} \right )^4, \ x\neq 0$

Answer:

Given the expression,

$\left(1 + \frac{x}{2} - \frac{2}{x} \right )^4, \ x\neq 0$

Binomial expansion of this expression is

$\\\left(1 + \frac{x}{2} - \frac{2}{x} \right )^4\\=\left ( \left (1 + \frac{x}{2} \right ) -\frac{2}{x}\right )^4=^4C_0\left ( 1+\frac{x}{2} \right )^4-^4C_1\left ( 1+\frac{x}{2} \right )^3\left ( \frac{2}{x} \right )+$$^4C_2\left ( 1+\frac{x}{2} \right )^2\left ( \frac{2}{x} \right )^2$$-^4C_3\left ( 1+\frac{x}{2} \right )\left ( \frac{2}{x} \right )^3+^4C_4\left ( \frac{2}{x} \right )^4$

$\Rightarrow \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^4}..........(1)$

Now Applying Binomial Theorem again,

$\left ( 1+\frac{x}{2} \right )^4=^4C_0(1)^4+^4C_1(1)^3\left ( \frac{x}{2} \right )+^4C_2(1)^2\left ( \frac{x}{2} \right )^2+^4C_3(1)\left ( \frac{x}{2} \right )^3$$+^4C_4\left ( \frac{x}{2} \right )^4$

$= 1+ 4\left ( \frac{x}{2} \right )+6\left ( \frac{x^2}{4} \right )+4\left ( \frac{x^3}{8} \right )+\frac{x^4}{16}$

$=1+2x+\frac{3x^2}{2}+\frac{x^3}{3}+\frac{x^4}{16}..............(2)$

And

$\left ( 1+\frac{x}{2} \right )^3=^3C_0(1)^3+^3C_1(1)^2\left ( \frac{x}{2} \right )+^3C_2(1)\left ( \frac{x}{2} \right )^2+^3C_3\left ( \frac{x}{2} \right )^3$

$\left ( 1+\frac{x}{2} \right )^3= 1+\frac{3x}{2}+\frac{3x^2}{4}+\frac{x^3}{8}..........(3)$

Now, From (1), (2) and (3) we get,

$\\\left(1 + \frac{x}{2} - \frac{2}{x} \right )^4=1+2x+\frac{3x^2}{2}+\frac{x^3}{8}+\frac{x^4}{16}-\frac{8}{x}\left ( 1+\frac{3x}{2}+\frac{3x^2}{4}+\frac{x^2}{8} \right )$$+\frac{8}{x^2}+\frac{24}{x}+6-\frac{32}{x^3}+\frac{16}{x^4}$

$\\\left(1 + \frac{x}{2} - \frac{2}{x} \right )^4=1+2x+\frac{3x^2}{2}+\frac{x^3}{8}+\frac{x^4}{16}-\frac{8}{x}-12-6x-x^2$$+\frac{8}{x^2}+\frac{24}{x}+6-\frac{32}{x^3}+\frac{16}{x^4}$

$\\\left(1 + \frac{x}{2} - \frac{2}{x} \right )^4=1+2x+\frac{3x^2}{2}+\frac{x^3}{8}+\frac{x^4}{16}-\frac{8}{x}\left ( 1+\frac{3x}{2}+\frac{3x^2}{4}+\frac{x^2}{8} \right )$$=\frac{16}{x}+\frac{8}{x^2}-\frac{32}{x^3}+\frac{16}{x^4}-4x+\frac{x^2}{2}+\frac{x^3}{2}+\frac{x^4}{16}-5$

Question:10 Find the expansion of $(3x^2 - 2ax +3a^2)^3$ using binomial theorem.

Answer:

Given $(3x^2 - 2ax +3a^2)^3$

By Binomial Theorem It can also be written as

$(3x^2 - 2ax +3a^2)^3=((3x^2 - 2ax) +3a^2)^3$

$= ^3C_0(3x^2-2ax)^3+^3C_1(3x^2-2ax)^2(3a^2)+^3C_2(3x^2-2ax)(3a^2)^2+^3C_3(3a^2)^3$

$= (3x^2-2ax)^3+3(3x^2-2ax)^2(3a^2)+3(3x^2-2ax)(3a^2)^2+(3a^2)^3$

$= (3x^2-2ax)^3+81a^2x^4-108a^3x^3+36a^4x^2+81a^4x^2-54a^5x+27a^6$

$= (3x^2-2ax)^3+81a^2x^4-108a^3x^3+117a^4x^2-54a^5x+27a^6...........(1)$

Now, Again By Binomial Theorem,

$(3x^2-2ax)^3=^3C_0(3x^2)^3-^3C_1(3x^2)^2(2ax)+^3C_2(3x^2)(2ax)^2-^3C_3(2ax)^3$

$(3x^2-2ax)^3= 27x^6-3(9x^4)(2ax)+3(3x^2)(4a^2x^2)-8a^2x^3$

$(3x^2-2ax)^3= 27x^6-54x^5+36a^2x^4-8a^3x^3............(2)$

From (1) and (2) we get,

$(3x^2-2ax+3a^2)^3=27x^6-54x^5+36a^2x^3+81a^2x^4-108a^3x^3+117a^4x^2$$-54a^5x+27a^6$

$(3x^2-2ax+3a^2)^3=27x^6-54x^5+117a^2x^3-116a^3x^3+117a^4x^2$$-54a^5x+27a^6$