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 :)
n = 1, 2n = 2 n = 2, 2n = 4 n = 3, 2n = 6 2, 4, 6, ..., 2n where n = 1, 2, 3, ... This is an arithmetic sequence, where the first term is 2 and the common difference is 2.
2n+4: 6,8,10......104........204
2n plus 4 = 2n + 4
The first three terms for the expression 2n-6 are obtained by substituting n with consecutive integers. When n=1, the expression evaluates to -4; when n=2, the expression evaluates to -2; and when n=3, the expression evaluates to 0. Therefore, the first three terms are -4, -2, and 0.
If you mean: 4n-2n = 4 then 2n = 4 and n = 2
2n-9=-4
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 :)
First we must collect and simplify 'like terms' 2n-6n-36. 2n-6n = -4n Therefore, -4n-36 Take out the greatest common factor. -4 (n+9) Ans = -4 (n+9)