Maths

$S_n={\displaystyle \sum _{r=1}^{n-1}r\left(n-r\right)}$

$=n\sum r-\sum r^2$

$=n\frac{\left(n-1\right)n}{2}-\frac{\left(n-1\right)n\left(2n-1\right)}{6}=\frac{n^3-n}{6}$

$S={\displaystyle \sum _{n=4}^{\infty }\left(\frac{2}{6}.\frac{n^3-n}{n!}-\frac{1}{\left(n-2\right)!}\right)=\frac{1}{3}}{\displaystyle \sum _{n=4}^{\infty }\frac{1}{\left(n-3\right)!}=\frac{e-1}{3}}$

a, b, c, d, e be 5 unknown

n = 7, mean = 8, variance = 16

sum of observations = 7 * 8 = 56

mean of 5 remaining observation = $\frac{56-8-6}{25}=\frac{42}{5}$

$16=\frac{\sum x_i^2}{7}-64$

$\sum x_i^2=560$

$\begin{array}{l}{a}^2+b^2+c^2+d^2+e^2=460\\ =460-{\left(\frac{42}{5}\right)}^2\end{array}$

$=\frac{536}{25}$                                                

$l_{n,m}={\displaystyle \underset{0}{\overset{\frac{1}{2}}{\int }}\frac{x^n}{x^m-1}dx,m,n\in N,n>m}$

$l_{6+i,3}-l_{3+i,3}$

$A=\frac{1}{2^5}\left[\begin{array}{ccc}\hfill \frac{1}{5}\hfill & \hfill \frac{1}{5}\hfill & \hfill \frac{1}{5}\hfill \\ \hfill 0\hfill & \hfill \frac{1}{12}\hfill & \hfill \frac{1}{12}\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \frac{1}{28}\hfill \end{array}\right]$ $=\frac{B}{32}$

$\left|A\right|={\left(\frac{1}{32}\right)}^3\left|B\right|=\frac{1}{105.2^{19}}$

$\frac{dy}{dx}=\frac{2^{x}y+2^y.2^x}{2^x+2^{x+y}lo{g}_{e}2}$

${\displaystyle \int \frac{1+2^{y}lo{g}_{e}2}{y+2^y}dy={\displaystyle \int dx}}$        

=> ln|y + 2^y| = x + c

y (0) = 0

ln |y + 2^y| = x

y = 1

x = ln 3

$x\in \left(1,2\right)$

Angle bisectors are

$\frac{x-2y-2z+1}{3}=\pm \frac{2x-3y-6z+1}{7}$              

->x – 5y + 4z + 4 = 0, (i)

3x – 23y – 32z + 10 = 0 . (ii)

As distance of a point (-1, 0,0) on x – 2y – 2z + 1 = 0

from (i) is greater than that form (ii)

(ii) is the acute angle bisector.

$l={\pi }^2{\displaystyle \underset{0}{\overset{2}{\int }}\left(sin\left(\frac{\pi x}{2}\right){\left(x-\left[x\right]\right)}^{\left[x\right]}\right)dx}$

$l={\pi }^2{\displaystyle \underset{0}{\overset{1}{\int }}sin\left(\frac{\pi x}{2}\right).x^{0}dx+{\pi }^2}{\displaystyle \underset{1}{\overset{2}{\int }}sin\left(\frac{\pi x}{2}\right){\left(x-1\right)}^{1}dx}$

$l={\pi }^2\left(\frac{2}{\pi }\right)+\frac{2{\pi }^2}{\pi }\left[1-0\right]+2\pi *\frac{2}{\pi }{\left[sin\frac{\pi x}{2}\right]}_1^2$

$\begin{array}{l}l=2\pi +2\pi +4\left(0-1\right)\\ l=4\pi -4\end{array}$

Required number =  - number of solution of x₁ + x₂ + x₃ = 15, where x₁ < x₂ < x₃

For x₂ = x₁ + a, a   $\ge 1$

->3x₁ + 2a + b = 15

Coefficient of x¹⁵ in

$\left(x^3+x^6+x^9+x^{12}+x^{15}\right)$

$\left(x^1+x^2+x^3+.....+x^{10}\right)=12$

Required number = 455 – 12 = 443

f (0) = 0, f (1) = 1, f (2) = 2

h (x) = f (x) – x has three roots

$\Rightarrow h\text{'}\left(x\right)=f\text{'}\left(x\right)-1$ , has at least two roots

$\Rightarrow h\text{'}\text{'}\left(x\right)=f\text{'}\text{'}\left(x\right)$ has at least one root.

$x+1-2lo{g}_2\left(3+2^x\right)+2lo{g}_4\left(10-2^{-x}\right)=0$

$x+1+lo{g}_2\left[\frac{10.2^x-1}{{\left(3+2^x\right)}^2}\right]-x=0$

${\left(2^x\right)}^2-14.2^x+11=0$

2x + y

$x_1+x_2=lo{g}_2\left(49-\frac{152}{4}\right)$

$x_1+x_2=lo{g}_{2}11$

$\frac{S_{10}}{S_P}=\frac{100}{P^2}$

$S_P=\frac{S_{10}.P^2}{100}\Rightarrow S_{11}=\frac{S_{10}.121}{100}$

$\frac{a_{11}}{a_{10}}=\frac{S_{11}-S_{10}}{S_{10}-S_9}=\frac{S_{10}.\frac{121}{100}-S_{10}}{S_{10}-\frac{S_{10}.81}{100}}$

= $\frac{21}{19}$