n+n-n-n-n+n-n-n squared to the 934892547857284579275348975297384579th power times 567896578239657824623786587346378 minus 36757544.545278789789375894789572356757583775389=n solve for n! the answer is 42
n2 + n = n(n + 1)
n^2 + n
N+n=0
n = 5
sal n battery n gelykstroom of wisselstroom voorsien
Gelykstroom
direkte stroom
Gelykstroom vloei in een rigting en bly konstant, terwyl wisselstroom wissel tussen positiewe en negatiewe rigtings teen 'n gereelde frekwensie. Wisselstroom word gebruik vir huishoudelike krag, terwyl gelykstroom gebruik word vir batterye en elektronika-toepassings.
n n n n n n n n.
n squared x n n x n x n = n cubed n x n = n squared n squared x n = n cubed
N - 5*N = 4*N N - 5*N = 4*N N - 5*N = 4*N N - 5*N = 4*N
(n*n)+n
jazz has been around for a billion years
Barbados \n . Botswana \n . Bulgaria \n . Cameroon \n . Colombia \n . Ethopia \n . Hondurus \n . Kiribati \n . Malaysia \n . Mongolia \n . Pakistan \n . Paraguay \n . Portugal \n . Slovakia \n .
n ,n ,n,n,,n ,,n,n
Assuming you mean the first n counting numbers then: let S{n} be the sum; then: S{n} = 1 + 2 + ... + (n-1) + n As addition is commutative, the sum can be reversed to give: S{n} = n + (n-1) + ... + 2 + 1 Now add the two versions together (term by term), giving: S{n} + S{n} = (1 + n) + (2 + (n-1)) + ... + ((n-1) + 2) + (n + 1) → 2S{n} = (n+1) + (n+1) + ... + (n+1) + (n+1) As there were originally n terms, this is (n+1) added n times, giving: 2S{n} = n(n+1) → S{n} = ½n(n+1) The sum of the first n counting numbers is ½n(n+1).