The sum of the first n consecutive even numbers can be found using S=n^2+n where n ≥ 2. What is the value of n when the sum is 210?

Step-by-step explanation:

Given the sum of the first n consecutive even numbers expressed as

S=n^2+n where n ≥ 2.

If S  = 210, then;

210 = n² + n

n² + n - 210  = 0

Factorize

n = -1±√1²-4(-210)/2

n = -1±√1+840/2

n = -1±√841/2

n = -1±29/2

n = -1+29/2 and -1-29/2

n = 28/2 and -30/2

n = 14 and -15

Since n ≥ 2, hence the required value of n is 14

based on context clues you gave i'm assuming the answer would be c.

answer: 2/4

step-by-step explanation: