+1 vote
58 views 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?

1. $116$
2. $210$
3. $263$
4. $478$

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## 2 Answers

+2 votes
according to the above function for <2> = 4+3+2+1= 10

& for <3> = 6+5+4+3+2+1=21

so the final value is 21*10 = 210

please verify it.
by Boss (13.1k points)
+2 votes
$\langle y \rangle = 2y+(2y-1)+(2y-2) + \dots +1$

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.
by Boss (45.8k points)

+1 vote
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