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).
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.
It's an algbraic expession in the form of: 2n+4
2n/3 = 4 2n = 3*4 2n = 12 n = 6 :)
'n' can be any number, but when applied to (2n + 4), n = 2.5
2n plus 4 = 2n + 4
If you mean: 4n-2n = 4 then 2n = 4 and n = 2
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).
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.
It's an algbraic expession in the form of: 2n+4
2n/3 = 4 2n = 3*4 2n = 12 n = 6 :)
'n' can be any number, but when applied to (2n + 4), n = 2.5
4
No. of subsets = 2n - 1 3 = 2n - 1 3 + 1 = 2n - 1 + 1 4 = 2n 4/2 = 2n/2 2/1 = 1n/1 2 = n n = 2elements
Un = 2n - 8 Un = 2n - 8 Un = 2n - 8 Un = 2n - 8
-3n+5n+4 = -4 2n = -4-4 2n = -8 n = -4