Maths Binomial Theorem

${\displaystyle \sum _{k=1}^{10}\left(2k-1\right).k{.}^{10}{C}_k}$

$=2\sum {k^2}^{10}{C}_k-\sum k{.}^{10}{C}_k$

$x=1\Rightarrow {\displaystyle \sum _{k=0}^{10}k}{.}^{10}{C}_k=10.2^9$

$x=1\Rightarrow \sum k^2{}^{10}{C}_k=90.2^8+10.2^9=2^8\left(90+201\right)=2^8.110$

S = 2⁹ . 110 – 10.2⁹ = 2⁹ . 100

S = 2⁹ . 100

$\frac{2^{11}.25}{2^1-1}$

$t_n=\frac{1}{2^n.3^{11-n}}={\left(\frac{1}{3}\right)}^{11}.{\left(\frac{3}{2}\right)}^n$

$S_{10}={\left(\frac{1}{3}\right)}^{11}.\frac{\left(\frac{3}{2}\right).\left({\left(\frac{3}{2}\right)}^{10}-1\right)}{\frac{3}{2}-1}$

$=\frac{1}{3^{10}}.\frac{3^{10}-2^{10}}{2^{10}}$

$=1{+}^{10}{C}_1.2^1{+}^{10}{C}_2.2^2+....{+}^{10}{C}_9.2^9$

k = 21 + 6m -> R = 3

S = 1 + 3 + 3² + 3³ + ….+ 3²⁰²¹   $=\frac{3^{2022}-1}{2}=\frac{1}{2}\left[a^{1011}-1\right]$

^{$=\frac{1}{2}\left[9^{1011}-1\right]=\frac{1}{2}\left[100k+10110-1-1\right]$}                          

= 50k₁ + 4

Remainder = 4

  $\left(1-x^2+3x^3\right){\left(\frac{5}{2}{x}^3-\frac{1}{5x^2}\right)}^{11},x\ne 0$

(r + 1)^th term for expansion of   ${\left(\frac{5}{2}{x}^3-\frac{1}{5x^2}\right)}^{11}$

  n – r = x₃

r = y   $0\le x,y\le 11shouldbenaturalnumber.$

= ¹¹C_y   $*{\left(\frac{5}{2}\right)}^x*{\left(-\frac{1}{5}\right)}^y*x^{\left(3x-2y\right)}$

for term to be independent of x, power of x should be zero.

(3x – 2y) = 0 (when 1 is multiplied).

3x + 3y = 33

y = $\frac{33}{5}\left(Nosolution\right)$  

3x – 2y + 2 = 0 where (-x²) is multiplied).

3x +  3y = 33

y = 7, x = 4

$\begin{array}{cc}\hfill 3x-2y+3=0\hfill & \hfill 3x+3y=33\hfill \end{array}\}$

$\therefore$   coefficient of term independent of x.

$\left(-1\right){*}^{11}\text{?}{C}_7*{\left(\frac{5}{2}\right)}^4*{\left(\frac{-1}{5}\right)}^7$

$=\frac{33}{200}$

$T_{r+1}{=}^{15}\text{?}{C}_r\cdot {\left(2x^{1/5}\right)}^{15-r}{\left(-x^{-1/5}\right)}^r$  

= ¹⁵C_r $\cdot 2^{15-r}\cdot x^{\frac{15-2r}{5}}\cdot {\left(-1\right)}^r$  

Coefficient of x^-1 ⇒ r = 10 ⇒ m = ¹⁵C₁₀ $\cdot 2^5$  

$x^{-3}\Rightarrow r=15\Rightarrow n{=}^{15}\text{?}{C}_{15}\cdot 2^0=1$               

now mn² = ¹⁵C_r   $\cdot 2^r$

⇒ r = 5

  $\underset{x\to 0}{lim}\frac{cos\left(sinx\right)-cosx}{x^4}=\underset{x\to 0}{lim}\frac{2sin\left(x+sinx\right).sin\left(\frac{x-sinx}{2}\right)}{x^4}$ $\underset{}{}\frac{}{{}^{}}\underset{}{}\frac{\frac{}{}}{{}^{}}$

$=\underset{x\to 0}{lim}2.\left(\frac{\left(\frac{x+sinx}{2}\right)\left(\frac{x-sinx}{2}\right)}{x^4}\right)$

= $\underset{x\to 0}{lim}\frac{1}{2}.\left(2-\frac{x^2}{3!}+\frac{x^4}{5!}........\right)\left(\frac{1}{3!}-\frac{x^2}{5!}-1\right)$ $\underset{}{}\frac{}{}\frac{{}^{}}{}\frac{{}^{}}{}\frac{}{}\frac{{}^{}}{}$

= $\frac{1}{6}$ $\frac{}{}$

System of equation can be written as

$\left(\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill 1\hfill \\ \hfill 5\hfill & \hfill -8\hfill & \hfill 9\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill a\hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{l}b\\ 3\\ -1\end{array}\right)$

$\left(\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill 1\hfill \\ \hfill 15\hfill & \hfill -24\hfill & \hfill 27\hfill \\ \hfill 6\hfill & \hfill 3\hfill & \hfill 3a\hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{l}b\\ 9\\ -3\end{array}\right)$

$R_3-2R_1,R_2-5R_1$

for no solution

3a + 9 = 0 but $\frac{3}{2}-\frac{9b}{2}\ne 0$

$\Rightarrow a=-3\Rightarrow b\ne \frac{1}{3}$