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

$S_{100}=100-\frac{\frac{2}{3}\left({\left(\frac{2}{3}\right)}^{100}-1\right)}{\frac{2}{3}-1}=100+2.\frac{\left(2^{100}-3^{100}\right)}{3^{100}}$

$=100-2+\frac{2^{101}}{3^{100}}=98+{\left(\frac{2}{3}\right)}^{100}.2<98+1$

$\left[\begin{array}{ccc}\hfill a_1\hfill & \hfill a_2\hfill & \hfill a_3\hfill \\ \hfill b_1\hfill & \hfill b_2\hfill & \hfill b_3\hfill \\ \hfill c_1\hfill & \hfill c_2\hfill & \hfill c_3\hfill \end{array}\right]a_1,b_1,c_{1}t\left\{-1,0,1\right\}$

$\underset{9times}{\underset{?}{a_1+a_2+a_3+....}}=5$

Total = 414

$\lambda =1+\mu$

$\begin{array}{l}\underset{\_}{2\lambda =3\lambda \mu }\\ \lambda =2\to S\left(2,4,6\right)\end{array}$

$\mu =1$

$\overrightarrow{n}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 5\hfill \end{array}\right|=\left(7,-2,-1\right)$

7x – 2y – z + d = 0

(2, 4, 6) Þ d = -14 + 8 + 6

d = 0

7x – 2y – z = 0

7 – 2 = 5

g (1) =?

$=f\left(f\left(f\left(1\right)\right)\right)+f\left(f\left(1\right)\right)$

$f\left(1\right)={\left(2\left(1-\frac{1}{2}\right)\left(2+1\right)\right)}^{\frac{1}{50}}=3^{\frac{1}{50}}$

$3^{\frac{1}{50}}+13^{\frac{1}{50}}>1^{\frac{1}{50}}$

3 > 1

$3^{\frac{1}{50}}>2$   $2>3^{\frac{1}{50}}>1$

$3<2^{50}$   $\left[3^{\frac{1}{50}}+1\right]=1+1=2$

${\alpha }_n=19^n-12^n$

$\frac{31{\alpha }_9-{\alpha }_{10}}{57.{\alpha }_2}=\frac{31\left(19^9-12^9\right)-\left(19^{10}-12^{10}\right)}{57\left(19^8-12^8\right)}$

$=\frac{19^9\left(31-19\right)-12^9\left(31-12\right)}{57\left(19^8-12^8\right)}$

$=\frac{19^9.12-12^9.19}{57\left(19^8-12^8\right)}=4$

$P\left(x_1,y_1\right)$

$Q\left(x_2,y_2\right)$

$x_1+x_2=\frac{r}{2},x_1{x}_2=\frac{P}{2}$

$y_1+y_2=s,y_1,y_2=-9$

$\left(x-x_1\right)\left(x-x_2\right)+\left(y-y_1\right)\left(y-y_2\right)=0$

$2\left(x^2+y^2\right)-rx-2sy+p-2q=0$

r = 11, s = 7, p – 2q = -22

The SITEEE exam syllabus is based on the CBSE class 11th and 12th syllabus.

${\left|2\left(\overrightarrow{a}\times \overrightarrow{b}\right)\right|}^2+4{\left(\overrightarrow{a}.\overrightarrow{b}\right)}^2$

= $4{\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^2$

=4 × 16 × 9 = 576

Symbiosis does not release the SITEEE syllabus officially. The syllabus is based on the classes 11th and 12th curriculum.

Yes. Candidates will have to declare preferences of choices for each round of NEET PG counselling after registration. It must be noted registration and choices can be declared online. At the same time, choices must be locked in the given deadline otherwise it gets locked automatically. Based on declared choices, seat availablity, NEET PG rank, seats will be allotted.

No. There is no negative marking in the SITEEE exam.

$x_1+x_2=6,13\to 6+6=12$

$\begin{array}{l}{x}_1+x_2=4,11,18\to 4+8+1=13\\ x_1+x_2=2,9,16\to 2+9+3=14\\ x_1+x_2=7,14\to 7+5=12\\ x_1+x_2=5,12\to 5+7=12\end{array}$

= 63

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

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

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

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

S = 2⁹ . 110 – 10.2⁹ = 2⁹ . 100

S = 2⁹ . 100

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

Candidates are selected for BHMS at Chandola Homoeopathic Medical College and Hospital based on their scores in the UPMT entrance exam. This is the Uttarakhand Pre-Medical Test and is comducted by the State Government of Uttarakhand at different centres. Candidates must appear for this exam followed by the counselling process to take admission in Chandola Homoeopathic Medical College and Hospital.

As per last year's analysis, VITEEE question paper difficulty ranges from easy to moderate. It can vary for each section (Physics/Chemistry/Mathematics/Biology/English/Aptitude). However, compared to JEE Main, the Vellore Institute of Technology Engineering Entrance Examination is considered easier.

Yes, candidates may be able to get admission in Chandola Homoeopathic Medical College and Hospital for BHMS with 60% in Class 12. The college has not specified the minimum required aggregate for admission, however considering 60% a decent score, candidates can apply for admission in Chandola Homoeopathic Medical College and Hospital.

Yes, Vellore Institute of Technology conducts the exam online for MPCEA and BPCEA streams. The test duration of CBT is 2 hours and 30 minutes, and is held at various test cities acrosss India and abroad.

The eligibility criteria of BHMS course at Chandola Homoeopathic Medical College and Hospital requires the candidates to pass Class 12 with requisite 5 subjects including Physics, Chemistry and Biology from a recognised board. They must appear for the UPMT admission test for admission in the BHMS course at Chandola Homoeopathic Medical College and Hospital.