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(1) $H{e}_2^{-}{\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^{1}B.O=\frac{3-2}{2}=\frac{1}{2}$ ${}_{}^{}{}_{}^{}{}_{}^{}{}_{}^{}\frac{}{}\frac{}{}$

(2) $H{e}_2^{+}{\sigma }_{1s}^2{\sigma }_{1s}^{*1}B.O=\frac{2-1}{2}=\frac{1}{2}$ ${}_{}^{}{}_{}^{}{}_{}^{}\frac{}{}\frac{}{}$

(3) $B{e}_2{\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}B.O=\frac{4-4}{2}=0$ ${}_{}{}_{}^{}{}_{}^{}{}_{}^{}{}_{}^{}\frac{}{}$

So Be2 does not exits

Given $l_{m,n}={\displaystyle {\int }_0^1{x}^{m-1}{\left(1-x\right)}^{n-1}dx............\left(i\right)}$  

put 1 - x = $t\{\begin{array}{l}x=0,t=1\\ x=1,t=0\end{array}$  

dx = -dt

From (i) $l_{m,n}={\displaystyle {\int }_1^0{\left(1-t\right)}^{m-1}.t^{n-1}\left(-dt\right)}$  

$l_{m,n}={\displaystyle {\int }_0^1{t}^{n-1}{\left(1-t\right)}^{m-1}dt={\displaystyle {\int }_0^1{x}^{n-1}{\left(1-x\right)}^{m-1}dx..........\left(ii\right)}}$                              

(i)   $l_{m,n}={\displaystyle {\int }_{\infty }^0\frac{1}{{\left(1+y\right)}^{m-1}}}{\left(1-\frac{1}{1+y}\right)}^{n-1}\left(\frac{-dy}{{\left(1+y\right)}^2}\right)={\displaystyle {\int }_0^{\infty }\frac{y^{n-1}}{{\left(1+y\right)}^{m+n}}dy}..........\left(iii\right)$

Similarly by (ii) $l_{m,n}={\displaystyle {\int }_0^{\infty }\frac{y^{m-1}}{{\left(y+1\right)}^{m+n}}dy..........\left(iv\right)}$  

Adding (iii) & (iv) $2l_{m,n}={\displaystyle {\int }_0^{\infty }\frac{y^{n-1}+y^{m+1}}{{\left(y+1\right)}^{m+n}}}$  

Putting   $\frac{1}{z}\{\begin{array}{l}y=1,z=1\\ y=\infty ,z=0\end{array}$ $\Rightarrow dy=\frac{-1}{z^2}dz$  

Hence  $l_{m,n}={\displaystyle {\int }_0^1\frac{x^{m-1}+x^{n-1}}{{\left(1+x\right)}^{m+n}}}$ dx = a lm, n

-> a = 1

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Kindly go throuigh the solution

Given   ${\displaystyle \sum _{i=1}^{18}\left(x_i-\alpha \right)=36i.e}{\displaystyle \sum _{i=1}^{18}{x}_i-18\alpha =36..........\left(i\right)}$

&       ${\displaystyle \sum _{i=1}^{18}{\left(x_i-\beta \right)}^2=90i.e{\displaystyle \sum _{i=1}^{18}{x_i}^2}}-2\beta {\displaystyle \sum _{}^{}{x}_i+18{\beta }^2=90.............\left(ii\right)}$      

(i) & (ii)   ${\displaystyle \sum _{i=1}^{18}{x_i}^2}=90-18\beta +36\beta \left(\alpha +2\right).............\left(iii\right)$

Now variance ${\sigma }^2=\frac{\sum {x_i}^2}{n}-{\left(\frac{\sum x_i}{n}\right)}^2$ = 1 given

->(a - b) (a - b + 4) = 0

Since $\alpha \ne \beta so\left|\alpha -\beta \right|=4$  

Angle of Ht – B – Ht > angle of Hb – B - Hb

Bond angle $\propto \mathrm{\%}S-Character\propto \frac{1}{\mathrm{\%}ofP-character}$ $\frac{}{}$

${}_{}$  $\angle H_t$ B Ht is more so % P-Character is less.

Given $f\left(x\right)=2x^5+5x^4+10x^3+10x^2+10x+10$

$f\left(-1\right)=3>0\&f\left(-2\right)=-34<0$           

So at least one root will lie in (-2, -1)

now $f\text{'}\left(x\right)=10x^4+20x^3+30x^2+20x+10$  

$=10\left[x^4+2x^3+3x^2+2x+1\right]$            

$=10x^2{\left(x+\frac{1}{x}+1\right)}^2>0\forall x\in R$           

So, f(x) be purely increasing function so exactly one root of f(x) that will lie in (-2, 1). Hence |a| = 2

Kindly consider the following figure

Kindly consider the following figure

$Given\left|z+5\right|\le 4..........\left(i\right)$

->Represent a circle ${\left(x+5\right)}^2+y^2\le 16$  

$=z\left(1+i\right)+\overline{z}\left(1-i\right)\ge -10..........\left(ii\right)$           

->Represent a line X – y $\ge -5$  

So max |z + 1|² = AQ²

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

$\begin{array}{l}=32+16\sqrt[]{2}\\ =\alpha +\beta \sqrt[]{2}\end{array}$        

Hence a + b = 48

Kindly consider the following figure

Yes, it is absolutely fine to get an idea from a sample online. However, do not make the mistake of copying the content in your cover letter for student visa. You can get a fair idea of the look and format of the letter, along with the details that should be included. However, be cautious of referring to an authentic source and always cross-check your draft before submission.  

In your cover letter for student visa, the authorities will determine your intent to come to Germany and your financial stability for the same. Hence, always provide a suitable justification for the programme you are applying for, personal and academic background, your plans after completing your degree, your intent to return and sources of funding. Other than this, your reasons to apply for a particular university and country-specific reasons are also essential.

The simple answer would be 'no'. Your motivation letter is not your German student visa cover letter. The requirements of both the documents are entirely different. Your SOP can be personal, focused more on your exposure and interest in the subject, while your cover letter will emphasize on your reasons to study and financial stability. There will some overlap of information in university-specific reasons and goals; however, avoid copying the content from your SOP.

The format of your cover letter for student visa application for Germany should be the same as any formal letter. Use a Times New Roman font and keep the margins at normal on each side of the page. Start by mentioning your name, address, email id, and date of submission, followed by a formal address to the authorities along with subject line. Please use professional language throughout your letter.

Kindly consider the following figure

Given $P_n={\alpha }^n+{\beta }^n,P_{n-1}=11\&P_{n+1}=29$  

$P_n={\alpha }^{n-2}.{\alpha }^2+{\beta }^{n-2}.{\beta }^2.........\left(i\right)$                             

Now quadratic equation having roots a & b will be x² – (a + b)x + ab = 0

i.e.         x² – x – 1 = 0,     put x = a and put x = b

So          a² = a + 1           & b² = b + 1

(i)  $P_n={\alpha }^{n-2}\left(\alpha +1\right)+{\beta }^{n-2}\left(\beta +1\right)$    

$P_n=P_{n-1}+P_{n-2}$                        

-> ${P_n}^2=234$

Kindly consider the following figure

18 = 3² * 2

For G.C.D to be 3. no. of four digits should be only multiple of 3, but not multiple of 9 & also should not be even.

As we know no. of the form                         

9 k -> 1000

9 k + 1 -> 1000

9 k + 2 -> 1000

9 k + 3 -> 1000  -> Total no. = 2000

9 k + 4 -> 1000

9 k + 5 -> 1000

9 k + 6 ->1000

9 k + 7 -> 1000

9 k + 8 -> 1000

In which half will be even & half be odd so Required no. = 1000