Sets

$\Delta =p!\left(p+1\right)!\left(p+2\right)!{\Delta }_1=2p!\left(p+1\right);\left(p+2\right)!=2{\left(p!\right)}^3.{\left(p+1\right)}^2.{\left(p+2\right)}^1\to \alpha +\beta =3+1=4$

${\Delta }_1=\left|\begin{array}{ccc}\hfill 1\hfill & \hfill p+1\hfill & \hfill \left(p+2\right)\left(p+1\right)\hfill \\ \hfill 1\hfill & \hfill p+2\hfill & \hfill \left(p+3\right)\left(p+2\right)\hfill \\ \hfill 1\hfill & \hfill p+3\hfill & \hfill \left(p+4\right)\left(p+3\right)\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill 0\hfill & \hfill -1\hfill & \hfill \left(p+2\right)\left(-2\right)\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill \left(p+3\right)\left(-2\right)\hfill \\ \hfill 1\hfill & \hfill p+3\hfill & \hfill \left(p+4\right)\left(p+3\right)\hfill \end{array}\right|$

= 2(p + 3) – 2 (p + 2) = 2

$S=2+\frac{6}{7}+\frac{12}{7^2}+\frac{20}{7^3}+\frac{30}{7^4}+........(i)$  

$\frac{1}{7}S=\frac{2}{7}+\frac{6}{7^2}+\frac{12}{7^3}+\frac{20}{7^4}+........(ii)$               

(i) - (ii)

$\frac{6}{7}S=2+\frac{4}{7}=\frac{6}{7^2}+\frac{8}{7^3}+...........(iii)$               

$\frac{6}{7^2}S=\frac{2}{7}+\frac{4}{7^2}+\frac{6}{7^3}+........(iv)$              

(iii) – (iv)

$=2+\left[\frac{1}{1-\frac{1}{7}}\right]=2\left(\frac{7}{6}\right)$

$\therefore 4S=8{\left(\frac{7}{6}\right)}^3={\left(\frac{7}{3}\right)}^3$      

Yes, Sandip University employs a normalization process for SU JEE (Sandip University Joint Entrance Examination) scores to ensure fairness across different question paper sets and candidate categories. This process adjusts scores to account for variations in difficulty levels among different exam versions, ensuring that no candidate is unfairly advantaged or disadvantaged due to the specific set of questions they received.

66. Given series is 1* 2* 3 + 2* 3 *4 + 3* 4 *5 + … to n term

a_n = (n^th term of A. P. 1, 2, 3, …) ´* (n^th terms of A. P. 2, 3, 4) *

i e, a = 1, d = 2- 1 = 1i e, a = 2, d = 3- 2 = 1

(nth term of A. P. 3, 4, 5)

i e, a = 3, d = 3 -4 = 1.

= [1 + (n -1) 1] *[2 + (n -1):1]* [3 + (n- 1) 1]

= (1 + n -1)*(2 + n -1)*(3 + n -1)

= n (n + 1)(n + 2)

= n(n² + 2n + n + 2)

=n³ + 2n² + 2n.

S_n = ∑n³ + 3 ∑n² + 2 ∑n

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

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

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

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

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

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

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

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

$=\frac{n(n+1)(n+2)(n+3)}{4}$

43. ${\displaystyle \sum _{}^{11}}\left(2+3^x\right)=\left(2+3^1\right)+\left(2+3^2\right)+.........+\left(2+3^{11}\right)$

$x=1$  [2+2+ ….+ (11lines)] + (31+ 32+……+311)

= $11*2+\frac{3\left(3^{11}-1\right)}{3-1}\left[?s_n=\frac{a\left(r^n-1\right)}{r-1},r\ge 1\right]$

= $22+\frac{3\left(3^{11}-1\right)}{2}$

62. Let H and E be set of students who known Hindi and English respectively.

Then, number of students who know Hindi = n (H) = 100

Number of students who know English = n (E) = 50

Number of students who know both English & Hindi = 25 = n (H$\cap$E)

As each of students knows either Hindi or English,

Total number of students in the group,

n (H$\cup$E) = n (H) + n (E) - n (H$\cap$E)

= 100 + 25

= 125,

61. Let T and C be sets of students taking tea and coffee.

Then, n (T) = 150, number of students taking tea

n (C) = 225, number of students taking coffee

n (T$\cap$C) = 100, number of students taking both tea and coffee.

So, Number of students taking either tea or coffee is.

n (T$\cup$C) = n (T) + n (C) n (T$\cap$C)

= 150 + 225 100

= 275

Number of students taking neither tea coffee

= Total number of students No of students taking either tea or coffee

= 600 275

= 325.

60. Let A = {x, y}

B = {y, z}

C = {x, z}

So, A$\cap$B = {x, y}$\cap$ {y, z} = {y}≠$?$

B$\cap$C = {y, z} $\cap$ {x, z} = {z}≠$?$

A$\cap$C = {x, y}$\cap$ {x, z} = {x}≠$?$

But A$\cap$B$\cap$C = (A$\cap$B)$\cap$ C

= {y} $\cap$ (x, z}

=$?$

58. Let A = {a}, B = {a, b}, C = {a, c}

So, A$\cap$B = {a} $\cap$ {a, b} = {a}

A$\cap$C = {a} $\cap$ {a, c} = {a}

i.e., A$\cap$B = A$\cap$C = {a}

But B ≠C. as b$\cancel{\in }$B but $b\cancel{\ne }C$ vice-versa