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The admission process for the BTech in Agricultural Engineering course can take 3 to 5 months. This is because every college adheres to their own admission guidelines that involve a lot of formalities, including entrance exams, college applications, results, merit cutoffs, college counseling, seat allotment, and fee payment. 

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Agricultural Engineering

105. Given,

Cost of machine =? 15625

depreciation rate = 20 % each year.

We have,

Depreciated value after 1st year =? 15625 - 20 % of 15625

=?  $15625-\frac{20}{100}\times$ ?15625

=?  $15625\left(1-\frac{20}{100}\right)$

=?  $15625\times \left(1-\frac{1}{5}\right)$

=?  $15625\times \frac{4}{5}$

Similarly,

Depreciated value after 2nd year =? 15625 $\times {\left(\frac{4}{5}\right)}^2$ and so on.

This, Depreciated value at end of 5 years

=? 15625 $\times {\left(\frac{4}{5}\right)}^5$

=? 15625 $\times \frac{1024}{3125}$

=? 5120

103. As the number of letters mailed forms a G.P. i.e., 4, 4², ……., 4⁸

We have,

a = 4

$r=\frac{4^2}{4}=4$

and n = 8

So, Total numbers of letters = 4 + 4³ + ……. + 4⁸

$=\frac{4\left(4^8-1\right)}{4-1}$

$=\frac{4\times (65536-1)}{3}$

$=\frac{4}{3}\times 65535$

$=\frac{4}{8}\times 21,845$

= 87380

As amount spent on one postage = 50 paise $=$ ?  $\frac{50}{100}$

So, for reqd. postage =?  $\frac{50}{100}\times 87380$

=? 43690

The students have good chances of getting into IIM Rohtak's IPM if they score well in the IPMAT Rohtak exam. A high IPMAT Rohtak score will indicate good academic performance. It will also help to get shortlisted for the interview round at IIM Rohtak. To get admission confirmation at IIM Rohtak, you also need to score well in the interview round. The IPMAT Rohtak score has a weight of 55 percent in the final merit list, and the Personal Interview has a weight of 15 percent. The academic performance in Class 10 and Class 12 has a weightage of 10 and 20 percent respectively. 

Yes, candidates in their last year of graduation can apply for the MMS course at TIMSR Mumbai. Candidates are required to apply for the course through provisional admission. Once the graduation results are announced, candidates can fill out the details accordingly. Moreover, an important thing to note is all admissions will be provisional until approved by the Directorate of Technical Education (DTE), Admission Regulating Authority (ARA), and University of Mumbai (UoM). If the admission is not approved by any of these authorities due to any discrepancy in documents or otherwise, the admission will stand cancelled

100. Given, $\frac{1\times 2^2+2\times 3^2+?+n{(n+1)}^2}{1^2\times 2+2^2+3+?+n^2(n+1)}$

For numerator,

a  (n^th term) = n (n + 1)² = n (n + 1)² = n (n² + 2n +1) = n³ + 2n² + n

So, $S_n={\displaystyle \sum _{}^{}{a}_n}={\displaystyle \sum _{}^{}{n}^3}+2{\displaystyle \sum _{}^{}{n}^2}+{\displaystyle \sum _{}^{}n}$

$=\frac{n^2(n+1){}^2}{4}+\frac{2\cdot (n+1)(2n+1)n}{6}+\frac{n(n+1)}{2}$

$=\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+\frac{2(2n+1)}{3}+1\right]$

$=\frac{n(n+1)}{2}\left[\frac{3n(n+1)+2\times 2(2n+1)+6}{6}\right]$

$=\frac{n(n+1)}{12}\left[3n^2+3n+8n+4n+6\right]$

$=\frac{n(n+1)}{12}\left[3n^2+11n+10\right]$

$=\frac{n(n+1)}{12}\left[3n^2+6n+5n+10\right]$

$=\frac{n(n+1)}{12}[(33n(n+2)+5(n+2)]$

$=\frac{n(n+1)(n+2)(3n+5)}{12}$

For denominator,

a_n (n^th term) = n² (n + 1) = n³ + n²

So,

$S_n={\displaystyle \sum _{}^{}{a}_n}={\displaystyle \sum _{}^{}{n}^3}+{\displaystyle \sum _{}^{}{n}^2}$

$=\frac{n^2(n+1){}^2}{4}+\frac{n(n+1)(2n+1)}{6}$

$=\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+\frac{(2n+1)}{3}\right]$

$=\frac{n(n+1)}{2}\left[\frac{3n(n+1)+2(2n+1)}{6}\right]$

$=\frac{n(n+1)}{2}\left[\frac{3n^2+3n+4n+2}{6}\right]$

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

$=\frac{n(n+1)}{2}\left[\frac{3n(n+2)+(n+2)}{6}\right]$

$=\frac{n(n+1)(n+2)(3n+1)}{12}]$

So, $\frac{1\times 2^2+2\times 3^2+?+n{(n+1)}^2}{1^2\times 2+2^2\times 3+?+n^2(n+1)}=\frac{\frac{n(n+1)(n+2)(3n+5)}{12}}{\frac{n(n+1)(n+2)(3n+1)}{12}}$

$=\frac{3n+5}{3n+1}$

The D. Y. Patil College of Engineering and Technology has offered decent packages to its BTech and M.Tech graduates. Check the table provided below for more clarity on overall placement packages offered between 2022-2024:

99. The given series is.

$\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}$ +…………. upto nth term,

The nth term is ,

$a_n=\frac{1^3+2^3+3^3+........+n^3}{1+3+5+........+n...terms}$ { A.P. of n terms and a = 1, d = 5-3 = 2}

$a_n=\frac{{\left[\frac{n(n+1)}{2}\right]}^2}{\frac{n}{2}[2\times 1+(n-1)2]}$

$a_n=\frac{\frac{n^2(n+1){}^2}{4}}{\frac{n}{2}[2+(n-1)2]}=\frac{\frac{n^2(n+1){}^2}{4}}{n[1+n-1]}$

$a_n=\frac{\frac{n^2(n+1){}^2}{4}}{n^2}=\frac{{(n+1)}^2}{4}$

$a_n=\frac{n^2+2x+1}{4}=\frac{n^2}{4}+\frac{n}{2}+\frac{1}{4}$

$\therefore Sn={\displaystyle \sum _{}^{}an}=\frac{1}{4}{\displaystyle \sum _{}^{}{x}^2}+\frac{1}{2}{\displaystyle \sum _{}^{}n+{\displaystyle \sum _{}^{}1}}$

$=\frac{1}{4}\times \frac{9(n+1)(2n+1)}{6}+\frac{1}{2}\times \frac{n(x+1)}{2}+\frac{1}{4}\times n$

$=\frac{n}{4}\left[\frac{(n+1)(2n+1)}{6}+(n+1)+1\right]$

$=\frac{n}{4}\left[\frac{2n^2+n(n+1)(2n+1)+6(x+1)+6}{6}\right]$

$=\frac{n}{4}\left[\frac{2n^2+n+2n+1+6n+6+6}{6}\right]$

$=\frac{n}{4}\left[\frac{2n^2+9n+13}{6}\right]$

$=\frac{n\left(2n^2+9n+13\right)}{24}$

The D. Y. Patil College of Engineering and Technology is known for its M.Tech and BTech course placements. Check out the table below to know about the overall placement packages recorded at D. Y. Patil College of Engineering and Technology placement in 2022-2024: 

98. Given,

S₁ = 1 + 2 + 3 + ………. + n $=n+\frac{(n+1)}{2}$

S₂ = 1² + 2² + 3² + ………… + n²  $=\frac{n(n+1)(2n+1)}{6}$

S₃ = 1³ + 2³ + 3³ + …………. + n³ =${\left[\frac{n(n+1)}{2}\right]}^2$

So, L.H.S. $=9S_2^2=9\times {\left[\frac{n(n+1)(2n+1)}{6}\right]}^2$

$=\frac{9\times n^2(n+1){}^2(2n+1){}^2}{36}$

$=\frac{n^2(n+1){}^2(2n+1){}^2}{4}$

R.H.S. $=S_3\left(1+8S_1\right)={\left[\frac{n(n+1)}{2}\right]}^2\left[1+8\times \frac{n(n+1)}{2}\right]$

$=\frac{n^2(n+1){}^2}{4}\times [1+4n(n+1)]$

$=\frac{n^2(n+1){}^2}{4}\left[1+4n^2+4n\right]$

$=\frac{n^2(n+1){}^2}{4}\left[{(2n)}^2+2(2n)1+1^2\right]$

$=\frac{n^2(n+1){}^2}{4}{(2n+1)}^2=$ L.H.S.

So, 9(S₂)² = S₃ ( 1 + 8S₁)