Exercise 5.1 - Chapter 5 Binomial Theorem Sequences and Series 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

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}
$