Exercise 5.1 - Chapter 5 - Binomial Theorem, Sequences and Series - 11th Maths Guide Samacheer Kalvi Solutions

You can Download the Exercise 5.1 - Chapter 5 - Binomial Theorem, Sequences and Series - 11th Maths Guide Samacheer Kalvi Solutions with expert answers for all chapters. Perfect for Tamil & English Medium students to revise the syllabus and score more marks in board exams. Download and share it with your friends

Share this to Friend on WhatsApp

Binomial Theorem, Sequences and Series
Ex 5.1
Question 1.
Expand (i) $\left(2 x^2-\frac{3}{x}\right)^3\left(\right.$ ii) $\left(2 x^2-3 \sqrt{1-x^2}\right)^4+\left(2 x^2+3 \sqrt{1-x^2}\right)^4$
Solution:
(i)
$
\begin{aligned}
\left(2 x^2-\frac{3}{x}\right)^3 & ={ }^3 \mathrm{C}_0\left(2 x^2\right)^3+{ }^3 \mathrm{C}_1\left(2 x^2\right)^2\left(-\frac{3}{x}\right)+{ }^3 \mathrm{C}_2\left(2 x^2\right)^1\left(-\frac{3}{x}\right)^2+{ }^3 \mathrm{C}_3\left(-\frac{3}{x}\right)^3 \\
& { }^3 \mathrm{C}_0={ }^3 \mathrm{C}_3=1 ;{ }^3 \mathrm{C}_1={ }^3 \mathrm{C}_2=3 \\
& =1(8)\left(x^6\right)+3\left(4 x^4\right)\left(-\frac{3}{x}\right)+3\left(2 x^2\right)\left(\frac{9}{x^2}\right)+1\left(-\frac{27}{x^3}\right) \\
& =8 x^6-\frac{36 x^4}{x}+\frac{54 x^2}{x^2}-\frac{27}{x^3} \\
& =8 x^6-36 x^3+54-\frac{27}{x^3}
\end{aligned}
$
(ii) Taking $2 x^2$ as $a$ and $3 \sqrt{1-x^2}$ as $b$ we have $(a-b)^4+(a+b)^4$
$
\begin{gathered}
\text { Now }(a-b)^4={ }^4 \mathrm{C}_0 a^4+{ }^4 \mathrm{C}_1 a^3(-b)+{ }^4 \mathrm{C}_2\left(a^2\right)(-b)^2+{ }^4 \mathrm{C}_3(a)(-b)^3+{ }^4 \mathrm{C}_4(-b)^4 \\
{ }^4 \mathrm{C}_0=1={ }^4 \mathrm{C}_4 ;{ }^4 \mathrm{C}_1=4={ }^4 \mathrm{C}_3 ;{ }^4 \mathrm{C}_2=\frac{4 \times 3}{2 \times 1}=6
\end{gathered}
$

$
=a^4-4 a^3 b+6 a^2 b^2-4 a b^3+b^4
$
Similarly $(a+b)^4=a^4+4 a^3 b+6 a^2 b^2+4 a b^3+b^4$
$
\therefore(a-b)^4+(a+b)^4=2\left[a^4+6 a^2 b^2+b^4\right]
$
Substituting the value of $a$ and $b$ we get
$
\begin{aligned}
& 2\left[\left(2 x^2\right)^4+6\left(2 x^2\right)^2\left(3 \sqrt{1-x^2}\right)^2+\left(3 \sqrt{1-x^2}\right)^4\right] \\
= & 2\left[16 x^8+6\left(4 x^4\right)\left(9\left(1-x^2\right)\right)+81\left(1-x^2\right)^2\right] \\
= & 2\left[16 x^8+216 x^4\left(1-x^2\right)+81\left(1-x^2\right)^2\right] \\
= & 2\left[16 x^8+216 x^4-216 x^6+81+81 x^4-162 x^2\right]
\end{aligned}
$

$
\begin{aligned}
& =2\left[16 x^8-216 x^6+297 x^4-162 x^2+81\right] \\
& =32 x^8-432 x^6+594 x^4-324 x^2+162
\end{aligned}
$
Question 2.
Compute
(i) $102^4$
(ii) $99^4$
(iii) $9^7$
Solution:
$
\begin{aligned}
& \text { (i) } 102^4=(100+2)^4=\left(10^2+2\right)^4 \\
& ={ }^4 \mathrm{C}_0\left(10^2\right)^4+{ }^4 \mathrm{C}_1\left(10^2\right)^3(2)+{ }^4 \mathrm{C}_2\left(10^2\right)^2(2)^2+{ }^4 \mathrm{C}_3\left(10^2\right)^1(2)^3+{ }^4 \mathrm{C}_4(2)^4 \\
& { }^4 \mathrm{C}_0=1={ }^4 \mathrm{C}_4 ;{ }^4 \mathrm{C}_1=4={ }^4 \mathrm{C}_3 ;{ }^4 \mathrm{C}_2=\frac{4 \times 3}{2 \times 1}=6 \\
& =1\left(10^8\right)+4\left(10^6\right)(2)+6\left(10^4\right)(4)+4\left(10^2\right)(8)+16 \\
& =100000000+8000000+240000+3200+16 \\
& =108243216 \\
& \text { (ii) } 994=(100-1)^4=\left(10^2-1\right)^4 \\
& ={ }^4 \mathrm{C}_0\left(10^2\right)^4+{ }^4 \mathrm{C}_1\left(10^2\right)^3(-1){ }^1+{ }^4 \mathrm{C}_2\left(10^2\right)^2(-1)^2+{ }^4 \mathrm{C}_3\left(10^2\right)^1(-1)^3+{ }^4 \mathrm{C}_4(-1)^4 \\
& =1\left(10^8\right)+4\left(10^6\right)(-1)+6\left(10^4\right)(1)+4\left(10^4\right)(-1)+(-1)^4 \\
& =100000000-4000000+60000-400+1 \\
& =100060001-4000400=96059601 \\
&
\end{aligned}
$

$
\begin{aligned}
& \text { (iii) } \begin{aligned}
9^7= & (10-1)^7 \\
& ={ }^7 \mathrm{C}_0\left(10^7\right)+{ }^7 \mathrm{C}_1\left(10^6\right)(-1)^1+{ }^7 \mathrm{C}_2(10)^5(-1)^2+{ }^7 \mathrm{C}_3(10)^4(-1)^3+{ }^7 \mathrm{C}_4(10)^3(-1)^4 \\
& +{ }^7 \mathrm{C}_5(10)^2(-1)^5+{ }^7 \mathrm{C}_6(10)^1(-1)^6+{ }^7 \mathrm{C}_7(-1)^7
\end{aligned} \\
&{ }^7 \mathrm{C}_0=1={ }^7 \mathrm{C}_7 ;{ }^7 \mathrm{C}_1=7={ }^7 \mathrm{C}_6 ;{ }^7 \mathrm{C}_2=\frac{7 \times 6}{2 \times 1}=21={ }^7 \mathrm{C}_5,{ }^7 \mathrm{C}_3=\frac{7 \times 6 \times 5}{3 \times 2 \times 1}=35={ }^7 \mathrm{C}_4 \\
&= 1(10000000)+7(1000000)(-1)+21(100000)(1)+35(10000)(-1)+35(1000)(1)+21(100)(-1)+7(10) \\
&(1)+1(-1) \\
&= 10000000-7000000+2100000-350000+35000-2100+70-1 \\
&= 12135070-7352101=4782969
\end{aligned}
$
Question 3.
Using binomial theorem, indicate which of the following two number is larger: $(1.01)^{1000000}, 10000$.
Solution:
$
(1.01)^{1000000}=(1+0.01)^{1000000}
$

$
\begin{aligned}
= & { }^{1000000} \mathrm{C}_0(1)^{1000000}+{ }^{1000000} \mathrm{C}_1(1)^{999999}(0.01)^1 \\
& +{ }^{1000000} \mathrm{C}_2(1)^{999998}(0.01)^2+{ }^{1000000} \mathrm{C}_3(1)^{999997}(0.01)^3+\ldots \ldots \ldots . \\
= & 1(1)+1000000 \times \frac{1}{10^2}+\frac{1000000 \times 999999}{2} \times \frac{1}{10000}+\ldots \ldots \ldots . \\
= & 1+10000+50 \times 999999+\ldots \ldots . .
\end{aligned}
$
which is $>10000$
So $(1.01)^{1000000}>10000$ (i.e.) $(1.01)^{1000000}$ is larger.
Question 4.
Find the coefficient of $x^{15}$ in $\left(x^2+\frac{1}{x^3}\right)^{10}$.
Solution:
General term $T_{r+1}={ }^{10} \mathrm{C}_r\left(x^2\right)^{10-r}\left(\frac{1}{x^3}\right)^r$.
$
\begin{aligned}
& ={ }^{10} \mathrm{C}_r x^{20-2 r} \frac{1}{x^{3 r}}={ }^{10} \mathrm{C}_r x^{20-2 r} \cdot x^{-3 r} \\
& ={ }^{10} \mathrm{C}_r x^{20-5 r}
\end{aligned}
$
To find coefficient of $\mathrm{x}^{15}$ we have to equate $\mathrm{x}$ power to 15
i.e. $20-5 r=15$
$
20-15=5 \mathrm{r} \Rightarrow 5 \mathrm{r}=5 \Rightarrow \mathrm{r}=5 / 5=1
$
So the coefficient of $x^{15}$ is ${ }^{10} \mathrm{C}_1=10$

Question 5.
Find the coefficient of $x^6$ and the coefficient of $x^2$ in $\left(x^2-\frac{1}{x^3}\right)^6$.
Solution:
$
\text { General term } \begin{aligned}
\mathrm{T}_{r+1} & ={ }^6 \mathrm{C}_r\left(x^2\right)^{6-r}\left(\frac{-1}{x^3}\right)^r . \\
& ={ }^6 \mathrm{C}_r x^{12-2 r}(-1)^r \frac{1}{x^{3 i}} \\
& ={ }^6 \mathrm{C}_r(-1)^r x^{12-2 r-3 r}={ }^6 \mathrm{C}_r(-1)^r x^{12-5 r}
\end{aligned}
$
To find coefficient of $x^6$
$12-5 r=6$
$12-6=5 r \Rightarrow 5 r=6 \Rightarrow r=6 / 5$ which is not an integer.
$\therefore$ There is no term involving $\mathrm{x}^6$.
To find coefficient of $x^2$
$
12-5 r=2
$

$
5 \mathrm{r}=12-2=10 \Rightarrow \mathrm{r}=2
$
So coefficient of $x^2$ is ${ }^6 \mathrm{C}_2(-1)^2=\frac{6 \times 5}{2 \times 1}(1)=15$
Question 6.
Find the coefficient of $x^4$ in the expansion of $\left(1+x^3\right)^{50}\left(x^2+\frac{1}{x}\right)^5$.
Solution:
$
\begin{aligned}
&\left(x^2+\frac{1}{x}\right)^5={ }^5 \mathrm{C}_0\left(x^2\right)^5+{ }^5 \mathrm{C}_1\left(x^2\right)^4\left(\frac{1}{x}\right)+{ }^5 \mathrm{C}_2\left(x^2\right)^3\left(\frac{1}{x}\right)^2 \\
&+{ }^5 \mathrm{C}_3\left(x^2\right)^2\left(\frac{1}{x}\right)^3+{ }^5 \mathrm{C}_4\left(x^2\right)\left(\frac{1}{x}\right)^4+{ }^5 \mathrm{C}_5\left(\frac{1}{x}\right)^5 \\
&{ }^5 \mathrm{C}_0=1={ }^5 \mathrm{C}_5 ;{ }^5 \mathrm{C}_1=5={ }^5 \mathrm{C}_4 ;{ }^5 \mathrm{C}_2=\frac{5 \times 4}{2 \times 1}=10={ }^5 \mathrm{C}_3 \\
&= x^{10}+5\left(x^8\right)\left(\frac{1}{x}\right)+10\left(x^6\right)\left(\frac{1}{x^2}\right)+10\left(x^4\right)\left(\frac{1}{x^3}\right)+5\left(x^2\right)\left(\frac{1}{x^4}\right)+\frac{1}{x^5} \\
&= x^{10}+5 x^7+10 x^4+10 x+\frac{5}{x^2}+\frac{1}{x^5} \\
&+\ldots .50 \mathrm{C}_{50}(1)^{\circ}\left(x^3\right)^{50} \\
&\left(1+x^3\right)^{50}= 50 \mathrm{C}_0(1)^{50}\left(x^3\right)^0+50 \mathrm{C}_1(1)^{49}\left(x^3\right)^1+50 \mathrm{C}_2(1)^{48}\left(x^3\right)^2+50 \mathrm{C}_3(1)^{47}\left(x^3\right)^3 \\
& 50 \mathrm{C}_0= 1=50 \mathrm{C}_{50}, 50 \mathrm{C}_1=50,50 \mathrm{C}_2=1225,{ }^{50} \mathrm{C}_3=19600 \\
&=(1)+(50) x^3+1225 x^6 7600 x^9 \ldots x^{150}
\end{aligned}
$
To find co-eff of $x^4$
$
\begin{aligned}
\left(1+x^3\right)^{50}\left(x^{2+\frac{1}{x}}\right)^5= & \left.1+50 x^3+1225 x^6+19600 x^9 \ldots x^{150}\right) \times \\
& \left(x^{10}+5 x^7+10 x^4+10 x+\frac{5}{x^2}+\frac{1}{x^5}\right)
\end{aligned}
$
when multiplying these terms, we get $x^4$ terms
$
\begin{aligned}
& =\left(1 \times 10 x^4\right)+\left(50 x^3 \times 10 x\right)+\left(1225 x^6 \times \frac{5}{x^2}\right)+\left(19600 x^9 \times \frac{1}{x^5}\right) \\
& =10 x^4+500 x^4+6125 x^4+19600 x^4 \\
& =26325 x^4
\end{aligned}
$
$\therefore$ The co-eff of $\mathrm{x}^4$ is 26325

Question 7.
Find the constant term of $\left(2 x^3-\frac{1}{3 x^2}\right)^5$.
Solution:
$
\text { General term } \begin{aligned}
\mathrm{T}_{r+1} & ={ }^5 \mathrm{C}_r\left(2 x^3\right)^{5-r}\left(\frac{-1}{3 x^2}\right)^r \\
& ={ }^5 \mathrm{C}_r 2^{5-r} x^{15-3 r}(-1)^r \frac{1}{3^r} \frac{1}{x^{2 r}} \\
& ={ }^5 \mathrm{C}_r \frac{2^{5-r}}{3^r}(-1)^r x^{15-3 r-2 r} \\
& ={ }^5 \mathrm{C}_r(-1)^r \frac{2^{5-r}}{3^r} x^{15-5 r}
\end{aligned}
$
To find the constant term
$
\begin{aligned}
15-5 r & =0 \Rightarrow 5 r=15 \Rightarrow r=3 \\
\therefore \text { Constant term } & ={ }^5 \mathrm{C}_3(-1)^3 \frac{2^{5-3}}{3^3} \\
& =\frac{5 \times 4 \times 3}{3 \times 2 \times 1}(-1) \frac{\left(2^2\right)}{3^3}=\frac{10(-1)(4)}{27}=\frac{-40}{27}
\end{aligned}
$

Question 8.
Find the last two digits of the number 300 .
Solution:
$
\begin{aligned}
& 3^{600}=3^{2 \times 300}=(9) 300=(10-1)^{300} \\
& ={ }^{300} \mathrm{C}_0(10)^{300}(-1)^0-{ }^{300} \mathrm{C}_1(10)^{299}(-1)^1+{ }^{300} \mathrm{C}_2(10)^{298}(-1)^2 \\
& +{ }^{300} \mathrm{C}_3(10)^{297}(-1)^3+\ldots \ldots . .+{ }^{300} \mathrm{C}_{299}(10)^1(-1)^{299}+{ }^{300} \mathrm{C}_{300}(-1)^{300} \\
& =10^{300}-300 \times 10^{299}+\ldots \ldots .-300(10)+1(1)
\end{aligned}
$
All the terms except last term are $\div$ by 100 . So the last two digits will be 01 .

Question 9.
If $n$ is a positive integer, show that, $9^{n+1}-8 n-9$ is always divisible by 64 .
Solution:
$
\begin{aligned}
9^{n+1}= & (1+8)^{n+1}=(n+1) \mathrm{C}_0(1)+(n+1) \mathrm{C}_1(1)^n(8)^1+{ }^{(n+1)} \mathrm{C}_2(8)^2 \\
& +(n+1) \mathrm{C}_3(8)^3+\ldots \ldots \ldots \\
= & 1+(n+1) 8+\frac{(n+1) n}{2 !}\left(8^2\right)+\ldots \ldots \\
\left(\text { i.e.) } 9^{n+1}=\right. & 1+8 n+8+\frac{(n+1) n}{2 !}\left(8^2\right)+\frac{(n+1)(n)(n-1)}{3 !}\left(8^3\right)+\ldots \ldots .
\end{aligned}
$
$
\left.\therefore 9^{\mathrm{n}+1}-8 \mathrm{n}-9=64 \text { [an integer }\right]
$
$\Rightarrow 9^{\mathrm{n}+1}-8 n-9$ is divisible by 64

Question 10.
If $n$ is an odd positive integer, prove that the coefficients of the middle terms in the expansion of $(x+y)^n$ are equal.
Solution:
Given $n$ is odd. So let $n=2 n+1$, where $n$ is an integer.
The expansion $(\mathrm{x}+\mathrm{y})^{\mathrm{n}}$ has $\mathrm{n}+1$ terms. $=2 \mathrm{n}+1+1=2(\mathrm{n}+1)$ terms which is an even number.
So the middle term are $\frac{t_2(n+1)}{2}=t_{n+1} t_{2(n+1)}=t_{n+1}$ and $t_{n+1+1}=t_{n+2}$
(i.e.) The middle terms are $\mathrm{t}_{n+1}$ and $\mathrm{t}_{n+2}$
$
\begin{aligned}
t_{n+1}={ }^{2 n+1} \mathrm{C}_n \text { and } t_{n+2}=t_{n+1+1}={ }^2{ }^{+1} \mathrm{C}_{n+1} \\
\text { Now } n+n+1=2 n+1 \\
\Rightarrow{ }^{2 n+1} \mathrm{C}_n={ }^{2 n+1} \mathrm{C}_{n+1}
\end{aligned}
$
$\Rightarrow$ The coefficient of the middle terms in $(\mathrm{x}+\mathrm{y})^{\mathrm{n}}$ are equal.

Question 11.
If $\mathrm{n}$ is a positive integer and $r$ is a non-negative integer, prove that the coefficients of $x^r$ and $x^{n-r}$ in the expansion of $(1+\mathrm{x})^{\mathrm{n}}$ are equal.
Solution:
In $(1+x)^n$ the general term is $\mathrm{t}_{r+1}={ }^n \mathrm{C}_r x^r$
$\therefore$ Coefficient of $x^r$ is ${ }^n \mathrm{C}_r$ and coefficient of $x^{n-r}$ is ${ }^n \mathrm{C}_{n-r}$.
Now ${ }^n \mathrm{C}_r=\frac{n !}{r !(n-r) !}$
And ${ }^n C_{n-r}=\frac{n !}{(n-r) !(n-n-r) !(n-n+r)}$
$
=\frac{n !}{(n-r) ! r !}
$
$(1)=(2)(\text { i.e. })^n \mathrm{C}_r={ }^n \mathrm{C}_{n-r}$
$\Rightarrow$ The coefficient of $x^r$ and $x^{n-r}$ are equal.
Question 12.
If $\mathrm{a}$ and $\mathrm{b}$ are distinct Integers, prove that $\mathrm{a}-\mathrm{b}$ is a factor of $\mathrm{a}^{\mathrm{n}}-\mathrm{b}^{\mathrm{n}}$, whenever $\mathrm{n}$ is a positive integer. [Hint: write $\mathrm{a}^{\mathrm{n}}=(\mathrm{a}-\mathrm{b}+\mathrm{b})^{\mathrm{n}}$ and expand]
Solution:

$
\Rightarrow{ }^n \mathrm{C}_{r-1}:{ }^n \mathrm{C}_r:{ }^n \mathrm{C}_{r+1}=1: 7: 42
$
(i.e.) $\frac{{ }^n C_{r-1}}{{ }^n C_r}=\frac{1}{7}$
and $\frac{{ }^n \mathrm{C}_r}{{ }^n \mathrm{C}_{r+1}}=\frac{7}{42}=\frac{1}{6}$
$
\begin{aligned}
\text { (1) } & \Rightarrow \frac{n !}{(r-1) !(n-\overline{r-1}) !} / \frac{n !}{r !(n-r) !}=\frac{1}{7} \\
\text { (i.e.) } & \frac{h !}{(r-1) !(n+1-r) !} \times \frac{r !(n-r) !}{h !}=\frac{1}{7} \\
\Rightarrow & \frac{r}{n+1-r}=\frac{1}{7} \Rightarrow 7 r=n+1-r \Rightarrow 8 r-n=1 \rightarrow(\mathrm{A})
\end{aligned}
$
$
\begin{aligned}
& \text { (2) } \Rightarrow \frac{n !}{r !(n-r) !} / \frac{n !}{(r+1) !(n-r+1) ![n-r-1]}=\frac{1}{6} \\
& \text { (i.e.) } \frac{h !}{r !(n-r) !} \times \frac{(r+1) !(n-r-1) !}{h !}=\frac{1}{6} \\
& \frac{(r+1)}{n-r}=\frac{1}{6} \\
& n-r=6 r+6 \\
& n-7 r=6 \quad \rightarrow(\mathrm{B}) \\
&
\end{aligned}
$
Solving (A) and (B)
$
\begin{array}{ll}
-n+8 r=1 & \rightarrow(\mathrm{A}) \\
n-7 r=6 & \rightarrow(\mathrm{B})
\end{array}
$
(A) $+(\mathrm{B}) \Rightarrow \quad r=7$
Substituting $r=7$ in (B)

$
\begin{aligned}
& n=6+7 \times 7 \\
& n=6+49=55
\end{aligned}
$
Question 15
. In the binomial coefficients of $\left(1+x^{\text {B }}\right)$, the coefficients of the $5^{\text {th }}, 6^{\text {th }}$ and 7 terms are in AP. Find all values of $n$.
Solution:
Coefficients of $\mathrm{T}_5, \mathrm{~T}_6, \mathrm{~T}_7$, are in A.P.
$
\therefore{ }^n C_4,{ }^n C_5,{ }^n C_6 \text {, are in A.P. }
$
$
\Rightarrow \frac{n(n-1)(n-2)(n-3)}{4.3 .2 .1}, \frac{n(n-1)(n-2)(n-3)(n-4)}{5.4 .3 .2 .1} \text {, }
$
$\frac{n(n-1)(n-2)(n-3)(n-4)(n-5)}{6.5 \cdot 4.3 .2 .1}$ are in A.P.
Multiplying each term by $\frac{4.3 .2 .1}{n(n-1)(n-2)(n-3)}$, we get $\Rightarrow 1, \frac{n-4}{5}, \frac{(n-4)(n-5)}{6.5}$ are in A.P.
$\Rightarrow 1, \frac{n-4}{5}, \frac{n^2-9 n+20}{30}$ are in A.P.
$
\begin{aligned}
& \therefore \frac{n-4}{5}-1=\frac{n^2-9 n+20}{30}-\frac{n-4}{5} \\
& \Rightarrow \frac{n-9}{5}=\frac{n^2-15 n+44}{30} \\
& \Rightarrow 6(n-9)=n^2-15 n+44
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow \frac{n-9}{5}=\frac{n^2-15 n+44}{30} \\
& \Rightarrow 6(n-9)=n^2-15 n+44 \\
& \Rightarrow n^2-21 n+98=0 \\
& \Rightarrow(\mathrm{n}-1)(\mathrm{n}-14)=0 \\
& \therefore \mathrm{n}=7,14
\end{aligned}
$
Question 16.
Prove that $\mathrm{C}_0^2+\mathrm{C}_1^2+\mathrm{C}_2^2+\ldots+C_n^2=\frac{2 n !}{(n !)^2}$.
Solution:
We know $\mathrm{C}_0+\mathrm{C}_1+\mathrm{C}_2+\ldots \ldots . \mathrm{C}_n=2^n$
and $\mathrm{C}_0 \mathrm{C}_r+\mathrm{C}_1 \mathrm{C}_{r+1}+\mathrm{C}_2 \mathrm{C}_{r+2}+\ldots \ldots . .+\mathrm{C}_{n-r} \mathrm{C}_n={ }^{2 n} \mathrm{C}_{n-r}$
Taking $r=0$ we get
$
\begin{aligned}
& \mathrm{C}_0 \mathrm{C}_0+\mathrm{C}_1 \mathrm{C}_1+\mathrm{C}_2 \mathrm{C}_2+\ldots \ldots . .+\mathrm{C}_n \mathrm{C}_n={ }^{2 n} \mathrm{C}_n \\
& \text { (i.e.) } \mathrm{C}_0^2+\mathrm{C}_1{ }^2+\mathrm{C}_2{ }^2+\ldots \ldots \ldots . .+\mathrm{C}_n{ }^2={ }^{2 n} \mathrm{C}_n=\frac{2 n !}{n !(2 n-n) !}=\frac{2 n !}{n ! n !}=\frac{2 n !}{(n !)^2}
\end{aligned}
$