I am assuming that 2n is an algebraic expression, and n is limited to positive integer values.
The first 4 multiples of 2n are 0 (2n*0), 2n (2n*1), 4n (2n*2), and 6n (2n*3). If you are looking for non-0 multiples, you would also include 8n (2n*4).
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You can't. 17 + 2n = 21 + 2n gives: 2n = 4 + 2n which gives 0=4 which is not possible so the sum is not solveable.
2n+4/2 term 1 = 3 term 2 = 4 term 3 = 5 term 4 = 6
It's an algbraic expession in the form of: 2n+4
Let 2n be the first integer, then the next on is 2n+2 and the one after is 2n+4 so adding all three we have 6n+6=108 or 6n=102 and we want 2n so divide we divide 6n by 3 and we have 2n=34, and 2n+2=36 and lastly 2n+4=38 so 34, 36, 38 are the even integers we seek.
2n/3 = 4 2n = 3*4 2n = 12 n = 6 :)