Ask & Answer: India's Largest Education Community

This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l}Let{T}_{r+1}betheconstanttermintheexpansionof{\left(x^2-\frac{1}{x^2}\right)}^{16}\\ \therefore T_{r+1}={}_{16}{C}_r{\left(x^2\right)}^{16-r}{\left(\frac{-1}{x^2}\right)}^r={}_{16}{C}_r{\left(x\right)}^{32-2r}{\left(-1\right)}^r.\frac{1}{x^{2r}}\\ ={\left(-1\right)}^r.{}_{16}{C}_r{\left(x\right)}^{32-2r-2r}={\left(-1\right)}^r.{}_{16}{C}_r{\left(x\right)}^{32-4r}\\ Forgettingconstantterm,32-4r=0\\ \Rightarrow r=8\\ \therefore T_{r+1}={\left(-1\right)}^8.{}_{16}{C}_8={}_{16}{C}_8\\ Hence,thevalueofthefilleris{}_{16}{C}_8.\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Givenseries\\ \Rightarrow \left(3^3-2^3\right)+\left(5^3-4^3\right)+\left(7^3-6^3\right)+\dots =\left(3^3+5^3+7^3+\dots \right)-\left(2^3+4^3+6^3+\dots \right)\\ =\left[3^3+5^3+7^3+\dots {\left(2n+1\right)}^3\right]-\left[2^3+4^3+6^3+\dots {\left(2n\right)}^3\right]\\ \therefore T_n={\left(2n+1\right)}^3-{\left(2n\right)}^3\\ =\left(2n+1-2n\right)\left[{\left(2n+1\right)}^2+\left(2n+1\right)\left(2n\right)+{\left(2n\right)}^2\right]\left[?a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\right]\\ =1.\left[4n^2+1+4n+4n^2+2n+4n^2\right]=12n^2+6n+1\\ \left(i\right)S_n={\displaystyle \sum _{}^{}{T}_n}=12{\displaystyle \sum _{}^{}{n}^2}+6{\displaystyle \sum _{}^{}n}+n\\ =12.\frac{n\left(n+1\right)\left(2n+1\right)}{6}+\frac{6n\left(n+1\right)}{2}+n\\ =2n\left(n+1\right)\left(2n+1\right)+3n\left(n+1\right)+n\\ =n\left[2\left(n+1\right)\left(2n+1\right)+3\left(n+1\right)+1\right]\\ =n\left[2\left(2n^2+3n+1\right)+3n+3+1\right]\\ =n\left[4n^2+6n+2+3n+3+1\right]=n\left[4n^2+9n+6\right]\\ =4n^3+9n^2+6n\\ \left(ii\right)S_{10}=4{\left(10\right)}^3+9{\left(10\right)}^2+6\left(10\right)=4\times 1000+9\times 100+60\\ =4000+960=4960\end{array}$

Yes, it is worth pursuing BPharm at Advance Group of Colleges as it has more than 20 years of experience in the education field. Advance Group of Colleges courses are offered to students at the UG and PG levels, mainly in the field of Pharmaceutical Sciences, Biotechnology, Paramedical Sciences, Education, and more. These courses are offered to students in full-time mode and at an affordable fee, making them accessible to every student. Further, the college offers a BPharm programme to students for a duration of 4 years. 

This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l}Theexpression {\left(x+y+z\right)}^{n}canbewrittena{\left[x+\left(y+z\right)\right]}^n\\ \therefore {\left[x+\left(y+z\right)\right]}^n={}_n{C}_0{x}^n{\left(y+z\right)}^0+{}_n{C}_1{\left(x\right)}^{n-1}\left(y+z\right)+{}_n{C}_2{\left(x\right)}^{n-2}{\left(y+z\right)}^2+\dots +{}_n{C}_n{\left(y+z\right)}^n\\ \therefore Numberofterms1+2+3+4+\dots +\left(n+1\right)=\frac{\left(n+1\right)\left(n+2\right)}{2}\\ Hence,thevalueofthefilleris\frac{\left(n+1\right)\left(n+2\right)}{2}\end{array}$

Adani Skill Development application forms are out for the academic year 2025-26 for the certificate course. To get the application form candidates are required to first fill out the form before the last date. The admission to the course is direct. Candidates can apply online through the official website of the college. Additionally, candidates must meet the eligibility requirements for the programme. 

BTech course is three specialisations is offered at IIAEIT. Standard engineering formats are used in the academic structure, with specific disciplines added in subsequent semesters. Students may need to exercise more initiative in the areas of placement possibilities and practical experience. While making a decision, it is critical to consider the course options, costs, and long-term job fit.

This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l}Heren=30whichiseven\\ \therefore thethe\; largestcoefficientin{\left(1+x\right)}^n={}_n{C}_{n/2}\\ So,thethe\; largestcoefficientin{\left(1+x\right)}^{30}={}_{30}{C}_{15}\\ Hence,thevalueofthefilleris{}_{30}{C}_{15}.\end{array}$

Advance Group of Colleges is a private college and was established in 2004. It is one of the leading colleges dedicated to providing quality education to students and holistic development. Advance Group of Colleges Kanpur, is affiliated with the Dr. APJ Abdul Kalam Technical University (AKTU) and is approved by the Pharmacy Council of India (PCI).

AIBM Ahmedabad offers BBA and MBA courses. The selection criteria is based on students merit. The application process at AIBM Ahmedabad conducted online. To secure a seat, students can check the following steps presented below:

1. Visit the official website and complete the online application process. 
2. Ensure eligibility and selection criteria for the UG and PG programme.
3. The selection criteria is based on merit/Gujarat Common Admission Services
4. Upon selected, pay the specified fee to confirm and secure your seat.

This is a Objective Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l}Givenexpressionis{\left(\frac{1}{x}+xsinx\right)}^{10}\\ Numberofterms=10+1=11\left(odd\right)\\ \therefore Middleterm{\left(\frac{n+1}{2}\right)}^{th}term=\frac{11+1}{2}=\frac{12}{2}=6^{th}term\\ T_6=T_{5+1}={}_{10}{C}_5{\left(\frac{1}{x}\right)}^{10-5}{\left(xsinx\right)}^5\\ \therefore {}_{10}{C}_5{\left(\frac{1}{x}\right)}^5.x^5.si{n}^{5}x=7\frac{7}{8}\Rightarrow {}_{10}{C}_5.si{n}^{5}x=\frac{63}{8}\\ \Rightarrow \frac{10\times 9\times 8\times 7\times 6}{5\times 4\times 3\times 2\times 1}.si{n}^{5}x=\frac{63}{8}\Rightarrow 252.si{n}^{5}x=\frac{63}{8}\\ \Rightarrow si{n}^{5}x=\frac{63}{8\times 252}\Rightarrow si{n}^{5}x=\frac{1}{32}\\ \Rightarrow si{n}^{5}x={\left(\frac{1}{2}\right)}^5\Rightarrow sinx=\frac{1}{2}\\ \Rightarrow sinx=sin\frac{\pi }{6}\\ \therefore x=n\pi +{\left(-1\right)}^n.\frac{\pi }{6}\\ Hence,thecorrectoptionis\left(c\right).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}WeknowthatacoinhasHeadandTail\left(H,T\right)\\ \therefore Whenacoinistossed6times,thenthe\\ Possibleoutcome=2^6=64\\ Hence,thecorrectoptionis\left(b\right).\end{array}$

This is a Objective Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l}Givenexpressionis{\left(1+x\right)}^{2n}\\ T_{r+1}={}_{2n}{C}_r{x}^r\\ \therefore Coefficientof{x}^n={}_{2n}{C}_n=A\left(Given\right)\\ Intheexpressionis{\left(1+x\right)}^{2n-1}\\ T_{r+1}={}_{2n-1}{C}_r{x}^r\\ \therefore Coefficientof{x}^n={}_{2n-1}{C}_{n-1}=B\left(Given\right)\\ So,\frac{A}{B}=\frac{{}_{2n}{C}_n}{{}_{2n-1}{C}_{n-1}}\\ =\frac{\frac{2n!}{n!\left(n\right)!}}{\frac{\left(2n-1\right)!}{\left(n-1\right)!\left(2n-1-n+1\right)!}}=\frac{\frac{2n!}{n!n!}}{\frac{\left(2n-1\right)!}{\left(n-1\right)!\left(n\right)!}}\\ =\frac{2n!}{n!n!}\times \frac{\left(n-1\right)!n!}{\left(2n-1\right)!}=\frac{2n\left(2n-1\right)!}{n!n\left(n-1\right)!}\times \frac{\left(n-1\right)!n!}{\left(2n-1\right)!}\\ =\frac{2}{1}=2:1\\ Hence,thecorrectoptionis\left(b\right).\end{array}$

The steps to follow for filling out the application form for Adani Skill Development courses are as follows: 

Step 1: Go to the official website of Adani Skill Development.

Step 2: Click on "Courses" to learn about all the courses on offer. 

Step 3: Candidates can tap on the course they wish to apply to and fill out the application for the course. 

Step 4: Once the application has been filled, candidates can pay the fee. 

This is an Objective Type Questions as classified in NCERT Exemplar