UGC NET CSE | June 2019 | Part 1 | Question: 22

Question

For all integers $y>1, \: \langle y \rangle = 2y+(2y-1)+(2y-2) + \dots +1$.

What is the value of $\langle 3 \rangle \times \langle 2 \rangle $? Where $\times$ is a multiplication operator?

• A. $116$
• B. $210$
• C. $263$
• D. $478$

Answer 1

Given Function $\langle y \rangle=2y+(2y-1)+(2y-2)+ \dots+1\to\qquad(i)$

put $y=2$ at equation $(i)$

$<2> = 4+3+2+1= 10$
&

Put $y=3$ equation $(i)$

$ <3> = 6+5+4+3+2+1=21$

So $\langle3\rangle\times\langle2\rangle=21*10=210$

Option $(B)$ is correct.

Answer 2

$ \langle y \rangle = 2y+(2y-1)+(2y-2) + \dots +1$

we can write like this

 $ \langle y \rangle = 1+2+3 +\dots +(2y-2) +(2y-1) +2y $

It is arithmetic progression whose first term is $a = 1$ and difference $d =1$ and last terms $l = 2y$

Sum of the AP series $S = \dfrac{n}{2}\left(a+ l \right)$

 Now, find the sum  $ \langle y \rangle = \dfrac{2y}{2}(1+2y)$

$ \implies \langle y \rangle = y(1+2y) $

$ \implies \langle y \rangle  =  y + 2y^{2}\rightarrow(1)$

Now find the value of  $ \langle 3 \rangle = 3 + 18 = 21$ and $ \langle 2 \rangle = 2 + 8 = 10$

The value of $ \langle 3 \rangle \times  \langle  2 \rangle  = 21 \times 10 = 210$

Option $(B)$ is the correct answer.

Answer 3

Option B is the correct answer

<y> is simply the sum of the natural numbers till 2y, starting from 1

i.e. <y> = 1+2+3+4+..........+2y

so <3> = 1+2+3+........+2*3 = 1+2+3+4+5+6 = 21

and <2> = 1+2+3+4 = 10

hence <3>*<2> = 21*10 = 210