Question

For all integers y > 1, (y)= 2y+(2y-1)+(2y-2)+...+1. What is the value of (3) x (2)? Where x is a multiplication operator?
(1) 116
(2) 210 (3) 263 (4)478

Answer 1

(3) x (2)  = 210   if (y)= 2y+(2y-1)+(2y-2)+...+1

Step-by-step explanation:

(y)=  2y + (2y-1) +  (2y-2) +.........................+ 1

this is an AP

with first Term  a = 2y

Common difference d = - 1

n = number of terms

Last term = First term + (n-1)d

2y  + (n-1)(-1) = 1

=> n-1  = 2y - 1

=> n = 2y

Sum =  (2y)(2y + 1)/2  = y(2y+1)

(y) = y(2y+1)

=> (3) = 3* 7 = 21

    (2) = 2 * 5 = 10

(3) x (2)  = 21 * 10  

=>  (3) x (2)  = 210  

option 2 is correct

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